💡Alright, so I have this problem about a sequence defined by some initial terms and a recursive formula. It's divided into two parts: the first part asks for which real numbers x and y the sequence becomes constant after some point, and the second part asks for the general term of the sequence. Hmm, okay, let me try to unpack this step by step.First, let me write down what's given. The sequence is {a_n}, with a_0 = x and a_1 = y. The recursive formula is a_{n+1} = (a_n * a_{n-1} + 1) / (a_n + a_{n-1}) for n = 1, 2, and so on. So, each term is defined based on the two previous terms.Part (1) is asking for which real numbers x and y does there exist a positive integer n_0 such that for all n >= n_0, a_n is constant. That means after some point, the sequence stops changing and just stays at the same value. So, I need to find the conditions on x and y that make this happen.Part (2) is more general: find the general term a_n. That probably involves finding a formula that can express a_n in terms of n, x, and y.Let me tackle part (1) first. If the sequence becomes constant after some n_0, then for n >= n_0, a_{n+1} = a_n = a_{n-1} = c, where c is some constant. So, plugging this into the recursive formula:c = (c * c + 1) / (c + c)Simplify the denominator: c + c = 2c. So,c = (c^2 + 1) / (2c)Multiply both sides by 2c:2c^2 = c^2 + 1Subtract c^2 from both sides:c^2 = 1So, c = 1 or c = -1.Therefore, if the sequence becomes constant, it must be that the constant is either 1 or -1. So, we need to find x and y such that after some n_0, all terms are 1 or all terms are -1.Now, let's think about how the sequence can reach this constant value. Suppose that starting from a_0 = x and a_1 = y, after some steps, the sequence becomes 1 or -1 and stays there. So, the terms before n_0 might not be constant, but from n_0 onwards, they are.Let me consider the case where the constant is 1. So, suppose that a_{n} = 1 for all n >= n_0. Then, for n = n_0 - 1, we have:a_{n_0} = (a_{n_0 - 1} * a_{n_0 - 2} + 1) / (a_{n_0 - 1} + a_{n_0 - 2}) = 1So,(a_{n_0 - 1} * a_{n_0 - 2} + 1) = (a_{n_0 - 1} + a_{n_0 - 2})Which simplifies to:a_{n_0 - 1} * a_{n_0 - 2} - a_{n_0 - 1} - a_{n_0 - 2} + 1 = 0Factor this:(a_{n_0 - 1} - 1)(a_{n_0 - 2} - 1) = 0So, either a_{n_0 - 1} = 1 or a_{n_0 - 2} = 1.Similarly, if a_{n_0 - 1} = 1, then looking back one step:a_{n_0 - 1} = (a_{n_0 - 2} * a_{n_0 - 3} + 1) / (a_{n_0 - 2} + a_{n_0 - 3}) = 1Which again gives:(a_{n_0 - 2} * a_{n_0 - 3} + 1) = (a_{n_0 - 2} + a_{n_0 - 3})So,(a_{n_0 - 2} - 1)(a_{n_0 - 3} - 1) = 0Thus, either a_{n_0 - 2} = 1 or a_{n_0 - 3} = 1.This seems to suggest that if the sequence becomes 1 at some point, then one of the previous two terms must also be 1, and so on, recursively. Therefore, unless x or y is 1, the sequence might not necessarily become 1. Wait, but if x or y is 1, does that guarantee the sequence becomes 1?Let me test with x = 1. Suppose x = 1, then a_0 = 1. Then, a_1 = y. Let's compute a_2:a_2 = (a_1 * a_0 + 1) / (a_1 + a_0) = (y * 1 + 1) / (y + 1) = (y + 1)/(y + 1) = 1So, a_2 = 1. Then, a_3 = (a_2 * a_1 + 1)/(a_2 + a_1) = (1 * y + 1)/(1 + y) = (y + 1)/(y + 1) = 1So, from a_2 onwards, it's 1. Therefore, if x = 1, regardless of y, the sequence becomes 1 starting from a_2.Similarly, if y = 1, let's see:a_0 = x, a_1 = 1.a_2 = (a_1 * a_0 + 1)/(a_1 + a_0) = (1 * x + 1)/(1 + x) = (x + 1)/(x + 1) = 1Then a_3 = (a_2 * a_1 + 1)/(a_2 + a_1) = (1 * 1 + 1)/(1 + 1) = 2/2 = 1So, again, from a_2 onwards, it's 1.What if neither x nor y is 1? Let's say x = 2 and y = 2.Compute a_2: (2*2 + 1)/(2 + 2) = 5/4 = 1.25a_3: (1.25*2 + 1)/(1.25 + 2) = (2.5 + 1)/3.25 = 3.5/3.25 ≈ 1.0769a_4: (1.0769*1.25 + 1)/(1.0769 + 1.25) ≈ (1.3461 + 1)/2.3269 ≈ 2.3461/2.3269 ≈ 1.008a_5: (1.008*1.0769 + 1)/(1.008 + 1.0769) ≈ (1.085 + 1)/2.0849 ≈ 2.085/2.0849 ≈ 1.00005So, it's approaching 1. So, in this case, even though x and y are not 1, the sequence is approaching 1. So, does it become constant at 1? Well, in the limit, yes, but does it ever actually become exactly 1? From the calculation, it seems to get closer and closer but might never exactly reach 1 unless it's already there.Wait, but in the problem statement, it's asking for there to exist a positive integer n_0 such that for all n >= n_0, a_n is constant. So, if the sequence approaches 1 but never actually becomes 1, then it doesn't satisfy the condition. So, in the case where x and y are not 1 or -1, the sequence might approach 1 but never actually become constant. Therefore, to have the sequence become constant, we need that at some point, a_n = 1 or a_n = -1, and then it stays there.From the earlier example, if x = 1 or y = 1, then the sequence becomes 1 from a_2 onwards. Similarly, if x = -1 or y = -1, let's see:Suppose x = -1, y = something.a_0 = -1, a_1 = y.a_2 = (a_1 * a_0 + 1)/(a_1 + a_0) = (-y + 1)/(y - 1) = (1 - y)/(y - 1) = -1So, a_2 = -1.Then a_3 = (a_2 * a_1 + 1)/(a_2 + a_1) = (-1 * y + 1)/(-1 + y) = (-y + 1)/(y - 1) = -1So, from a_2 onwards, it's -1.Similarly, if y = -1:a_0 = x, a_1 = -1.a_2 = (a_1 * a_0 + 1)/(a_1 + a_0) = (-x + 1)/(-1 + x) = (1 - x)/(x - 1) = -1Then a_3 = (a_2 * a_1 + 1)/(a_2 + a_1) = (-1 * (-1) + 1)/(-1 + (-1)) = (1 + 1)/(-2) = 2/-2 = -1So, again, from a_2 onwards, it's -1.Therefore, if either x or y is 1 or -1, the sequence becomes constant at 1 or -1 respectively from a_2 onwards.But what if neither x nor y is 1 or -1? For example, x = 2, y = 3.Compute a_2: (2*3 + 1)/(2 + 3) = 7/5 = 1.4a_3: (1.4*3 + 1)/(1.4 + 3) = (4.2 + 1)/4.4 = 5.2/4.4 ≈ 1.1818a_4: (1.1818*1.4 + 1)/(1.1818 + 1.4) ≈ (1.6545 + 1)/2.5818 ≈ 2.6545/2.5818 ≈ 1.028a_5: (1.028*1.1818 + 1)/(1.028 + 1.1818) ≈ (1.213 + 1)/2.2098 ≈ 2.213/2.2098 ≈ 1.0014So, it's approaching 1 but not exactly reaching it. So, in this case, the sequence doesn't become constant; it just approaches 1 asymptotically.Therefore, to have the sequence become constant, we need that either x or y is 1 or -1. Because only then does the sequence reach the constant value in finite steps.Wait, but what if both x and y are 1? Then, a_2 = (1*1 + 1)/(1 + 1) = 2/2 = 1, and it stays 1. Similarly, if both are -1, a_2 = ((-1)*(-1) + 1)/(-1 + (-1)) = (1 + 1)/(-2) = 2/-2 = -1, and it stays -1.What if x = 1 and y = -1?a_0 = 1, a_1 = -1.a_2 = (a_1 * a_0 + 1)/(a_1 + a_0) = (-1*1 + 1)/(-1 + 1) = ( -1 + 1)/0 = 0/0, which is undefined.So, in this case, the sequence is undefined at a_2. Therefore, x and y cannot be 1 and -1 simultaneously because it leads to division by zero.Similarly, if x = -1 and y = 1, same problem:a_0 = -1, a_1 = 1.a_2 = (1*(-1) + 1)/(1 + (-1)) = (-1 + 1)/0 = 0/0, undefined.Therefore, x and y cannot be 1 and -1 simultaneously.So, to summarize part (1):The sequence becomes constant after some n_0 if and only if either x or y is 1 or -1, but not both 1 and -1 at the same time, as that would cause division by zero.Wait, but actually, if x = 1 and y = 1, it's fine, as we saw earlier. Similarly, x = -1 and y = -1 is fine. The problem is only when one is 1 and the other is -1.So, more precisely, the conditions are:Either x = 1 or y = 1, but not both x = 1 and y = -1,Or x = -1 or y = -1, but not both x = -1 and y = 1.But actually, if x = 1 and y = -1, it's undefined, as we saw. Similarly, x = -1 and y = 1 is undefined.Therefore, the conditions are:Either x = 1 and y ≠ -1,Or y = 1 and x ≠ -1,Or x = -1 and y ≠ 1,Or y = -1 and x ≠ 1.Because if x = 1 and y = -1, or x = -1 and y = 1, the sequence becomes undefined at a_2.So, to avoid division by zero, we need that if x = 1, then y ≠ -1,If y = 1, then x ≠ -1,If x = -1, then y ≠ 1,If y = -1, then x ≠ -1.Wait, no, if x = -1, y can be anything except 1, because if y = 1, then a_2 would be undefined.Similarly, if y = -1, x can be anything except 1.So, putting it all together, the real numbers x and y must satisfy:Either x = 1 and y ≠ -1,Or y = 1 and x ≠ -1,Or x = -1 and y ≠ 1,Or y = -1 and x ≠ 1.In all these cases, the sequence becomes constant at 1 or -1 starting from a_2.Therefore, the answer to part (1) is that x and y must be such that either x or y is 1 or -1, but not both 1 and -1 simultaneously, to avoid division by zero.Now, moving on to part (2): Find the general term a_n.This seems more involved. The recursive formula is a_{n+1} = (a_n * a_{n-1} + 1)/(a_n + a_{n-1}).This looks similar to some known sequences, but I'm not sure exactly which one. Maybe it's related to continued fractions or something else.Let me try to manipulate the recursive formula to see if I can find a pattern or a substitution that simplifies it.Let me denote b_n = (a_n - 1)/(a_n + 1). This substitution often helps in sequences involving terms like (a - 1)/(a + 1), which can sometimes turn multiplicative relations into additive ones.So, let's compute b_{n+1} in terms of b_n and b_{n-1}.First, express a_n in terms of b_n:From b_n = (a_n - 1)/(a_n + 1),Solving for a_n:b_n (a_n + 1) = a_n - 1b_n a_n + b_n = a_n - 1b_n a_n - a_n = -1 - b_na_n (b_n - 1) = - (1 + b_n)Therefore,a_n = -(1 + b_n)/(b_n - 1) = (1 + b_n)/(1 - b_n)So, a_n = (1 + b_n)/(1 - b_n)Now, let's compute a_{n+1}:a_{n+1} = (a_n a_{n-1} + 1)/(a_n + a_{n-1})Express a_n and a_{n-1} in terms of b_n and b_{n-1}:a_n = (1 + b_n)/(1 - b_n)a_{n-1} = (1 + b_{n-1})/(1 - b_{n-1})So,a_n a_{n-1} = [(1 + b_n)/(1 - b_n)] * [(1 + b_{n-1})/(1 - b_{n-1})]Similarly,a_n + a_{n-1} = [(1 + b_n)/(1 - b_n)] + [(1 + b_{n-1})/(1 - b_{n-1})]Let me compute a_n a_{n-1} + 1:= [(1 + b_n)(1 + b_{n-1})]/[(1 - b_n)(1 - b_{n-1})] + 1= [ (1 + b_n + b_{n-1} + b_n b_{n-1}) + (1 - b_n - b_{n-1} + b_n b_{n-1}) ] / [(1 - b_n)(1 - b_{n-1})]Wait, no, that's not correct. Let me compute it step by step.First, compute a_n a_{n-1}:= [(1 + b_n)(1 + b_{n-1})]/[(1 - b_n)(1 - b_{n-1})]Then, a_n a_{n-1} + 1 = [(1 + b_n)(1 + b_{n-1})]/[(1 - b_n)(1 - b_{n-1})] + 1To combine these, we need a common denominator:= [(1 + b_n)(1 + b_{n-1}) + (1 - b_n)(1 - b_{n-1})] / [(1 - b_n)(1 - b_{n-1})]Expand the numerator:= [ (1 + b_n + b_{n-1} + b_n b_{n-1}) + (1 - b_n - b_{n-1} + b_n b_{n-1}) ] / [(1 - b_n)(1 - b_{n-1})]Simplify numerator:= [1 + b_n + b_{n-1} + b_n b_{n-1} + 1 - b_n - b_{n-1} + b_n b_{n-1}] / denominator= [2 + 2 b_n b_{n-1}] / [(1 - b_n)(1 - b_{n-1})]So, a_n a_{n-1} + 1 = 2(1 + b_n b_{n-1}) / [(1 - b_n)(1 - b_{n-1})]Now, compute a_n + a_{n-1}:= [(1 + b_n)/(1 - b_n)] + [(1 + b_{n-1})/(1 - b_{n-1})]= [ (1 + b_n)(1 - b_{n-1}) + (1 + b_{n-1})(1 - b_n) ] / [(1 - b_n)(1 - b_{n-1})]Expand numerator:= [ (1 - b_{n-1} + b_n - b_n b_{n-1}) + (1 - b_n + b_{n-1} - b_n b_{n-1}) ] / denominatorSimplify numerator:= [1 - b_{n-1} + b_n - b_n b_{n-1} + 1 - b_n + b_{n-1} - b_n b_{n-1}] / denominator= [2 - 2 b_n b_{n-1}] / [(1 - b_n)(1 - b_{n-1})]So, a_n + a_{n-1} = 2(1 - b_n b_{n-1}) / [(1 - b_n)(1 - b_{n-1})]Therefore, a_{n+1} = (a_n a_{n-1} + 1)/(a_n + a_{n-1}) = [2(1 + b_n b_{n-1}) / denominator] / [2(1 - b_n b_{n-1}) / denominator] = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})So, a_{n+1} = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})But we also have a_{n+1} expressed in terms of b_{n+1}:a_{n+1} = (1 + b_{n+1}) / (1 - b_{n+1})Therefore,(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})Cross-multiplying:(1 + b_{n+1})(1 - b_n b_{n-1}) = (1 + b_n b_{n-1})(1 - b_{n+1})Expand both sides:Left side: 1 - b_n b_{n-1} + b_{n+1} - b_{n+1} b_n b_{n-1}Right side: 1 - b_{n+1} + b_n b_{n-1} - b_n b_{n-1} b_{n+1}Set them equal:1 - b_n b_{n-1} + b_{n+1} - b_{n+1} b_n b_{n-1} = 1 - b_{n+1} + b_n b_{n-1} - b_n b_{n-1} b_{n+1}Subtract 1 from both sides:- b_n b_{n-1} + b_{n+1} - b_{n+1} b_n b_{n-1} = - b_{n+1} + b_n b_{n-1} - b_n b_{n-1} b_{n+1}Bring all terms to the left side:- b_n b_{n-1} + b_{n+1} - b_{n+1} b_n b_{n-1} + b_{n+1} - b_n b_{n-1} + b_n b_{n-1} b_{n+1} = 0Wait, this seems messy. Maybe there's a simpler way.Alternatively, from the equality:(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})Let me denote c_n = b_n b_{n-1}, then:(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + c_n) / (1 - c_n)Cross-multiplying:(1 + b_{n+1})(1 - c_n) = (1 + c_n)(1 - b_{n+1})Expand:1 - c_n + b_{n+1} - b_{n+1} c_n = 1 - b_{n+1} + c_n - b_{n+1} c_nCancel terms:1 - c_n + b_{n+1} - b_{n+1} c_n = 1 - b_{n+1} + c_n - b_{n+1} c_nSubtract 1 from both sides:- c_n + b_{n+1} - b_{n+1} c_n = - b_{n+1} + c_n - b_{n+1} c_nBring all terms to the left:- c_n + b_{n+1} - b_{n+1} c_n + b_{n+1} - c_n + b_{n+1} c_n = 0Wait, this is getting too convoluted. Maybe I made a mistake in the substitution.Alternatively, let's consider the cross-ratio or some other property. Alternatively, perhaps the sequence b_n satisfies a multiplicative recurrence.Wait, from the earlier step:a_{n+1} = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})But a_{n+1} is also equal to (1 + b_{n+1}) / (1 - b_{n+1})Therefore,(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + b_n b_{n-1}) / (1 - b_n b_{n-1})Let me denote this as:(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + c_n) / (1 - c_n), where c_n = b_n b_{n-1}This suggests that b_{n+1} = c_n, because if (1 + x)/(1 - x) = (1 + y)/(1 - y), then x = y.Wait, let's test this:If (1 + x)/(1 - x) = (1 + y)/(1 - y), then cross-multiplying:(1 + x)(1 - y) = (1 + y)(1 - x)1 - y + x - xy = 1 - x + y - xySimplify:1 - y + x - xy = 1 - x + y - xyCancel 1 and -xy:- y + x = -x + yBring all terms to one side:- y + x + x - y = 02x - 2y = 0 => x = yTherefore, if (1 + x)/(1 - x) = (1 + y)/(1 - y), then x = y.Therefore, from our earlier equation:(1 + b_{n+1}) / (1 - b_{n+1}) = (1 + c_n) / (1 - c_n)This implies that b_{n+1} = c_n = b_n b_{n-1}So, we have:b_{n+1} = b_n b_{n-1}That's a multiplicative recurrence relation! So, b_{n+1} = b_n b_{n-1}This is a second-order multiplicative recurrence, which is similar to the Fibonacci recurrence but multiplicative.So, the sequence {b_n} satisfies b_{n+1} = b_n b_{n-1}Given that, we can try to find a general formula for b_n.This recurrence is known to generate terms that grow exponentially or decay exponentially, depending on the initial conditions.Given that, let's denote the initial terms:b_0 = (a_0 - 1)/(a_0 + 1) = (x - 1)/(x + 1)b_1 = (a_1 - 1)/(a_1 + 1) = (y - 1)/(y + 1)Then, b_2 = b_1 b_0b_3 = b_2 b_1 = b_1^2 b_0b_4 = b_3 b_2 = b_1^3 b_0^2b_5 = b_4 b_3 = b_1^5 b_0^3Wait, this seems to follow the Fibonacci sequence in exponents.Let me see:b_0 = b_0b_1 = b_1b_2 = b_1 b_0b_3 = b_2 b_1 = b_1^2 b_0b_4 = b_3 b_2 = b_1^3 b_0^2b_5 = b_4 b_3 = b_1^5 b_0^3b_6 = b_5 b_4 = b_1^8 b_0^5So, the exponents of b_1 and b_0 follow the Fibonacci sequence.Indeed, the exponents are Fibonacci numbers:For b_n, the exponent of b_1 is F_{n-1}, and the exponent of b_0 is F_{n-2}, where F_n is the nth Fibonacci number.Recall that Fibonacci numbers are defined as F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, etc.Wait, but in our case, for b_2, which is b_1 b_0, the exponents are F_1 = 1 and F_0 = 0? Wait, no, F_0 is 0, but in our case, b_2 = b_1^1 b_0^1, which would correspond to F_1 = 1 and F_0 = 0, but that doesn't fit.Wait, perhaps it's better to index the Fibonacci numbers differently.Let me define F_0 = 1, F_1 = 1, F_2 = 2, F_3 = 3, F_4 = 5, etc. So, F_n = F_{n-1} + F_{n-2} with F_0 = 1, F_1 = 1.Then, for b_n:b_2 = b_1 b_0 = b_1^{F_1} b_0^{F_0} = b_1^1 b_0^1b_3 = b_2 b_1 = (b_1 b_0) b_1 = b_1^2 b_0^1 = b_1^{F_2} b_0^{F_1} = b_1^2 b_0^1b_4 = b_3 b_2 = (b_1^2 b_0) (b_1 b_0) = b_1^3 b_0^2 = b_1^{F_3} b_0^{F_2} = b_1^3 b_0^2b_5 = b_4 b_3 = (b_1^3 b_0^2) (b_1^2 b_0) = b_1^5 b_0^3 = b_1^{F_4} b_0^{F_3} = b_1^5 b_0^3Yes, this seems to fit. So, in general,b_n = b_1^{F_{n-1}} b_0^{F_{n-2}}Where F_n is the nth Fibonacci number with F_0 = 1, F_1 = 1, etc.Therefore, b_n = [(y - 1)/(y + 1)]^{F_{n-1}} * [(x - 1)/(x + 1)]^{F_{n-2}}Now, recall that a_n = (1 + b_n)/(1 - b_n)Therefore,a_n = [1 + b_n] / [1 - b_n] = [1 + b_1^{F_{n-1}} b_0^{F_{n-2}}] / [1 - b_1^{F_{n-1}} b_0^{F_{n-2}}]Substituting b_0 and b_1:= [1 + [(y - 1)/(y + 1)]^{F_{n-1}} [(x - 1)/(x + 1)]^{F_{n-2}}] / [1 - [(y - 1)/(y + 1)]^{F_{n-1}} [(x - 1)/(x + 1)]^{F_{n-2}}]This is a general formula for a_n in terms of x, y, and Fibonacci numbers.Alternatively, we can write it as:a_n = [ ( (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} + (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ) ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} - (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ) ]Because:[(y - 1)/(y + 1)]^{F_{n-1}} = (y - 1)^{F_{n-1}} / (y + 1)^{F_{n-1}}Similarly for [(x - 1)/(x + 1)]^{F_{n-2}} = (x - 1)^{F_{n-2}} / (x + 1)^{F_{n-2}}Therefore, the numerator becomes:1 + [ (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} ]= [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} + (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} ]Similarly, the denominator becomes:1 - [ (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} ]= [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} - (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} ]Therefore, when we take the ratio, the denominators cancel out, leaving:a_n = [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} + (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ] / [ (y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} - (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}} ]This is the general term for a_n.To make it more symmetric, we can write it as:a_n = [ (x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} + (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}} ] / [ (x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} - (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}} ]Which is the same as:a_n = frac{(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} + (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}}{(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} - (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}}This formula gives the general term a_n in terms of x, y, and the Fibonacci numbers F_{n-1} and F_{n-2}.To verify this formula, let's test it with n = 0, 1, 2.For n = 0:a_0 = xUsing the formula:F_{-2} and F_{-1} are not defined in the standard Fibonacci sequence. So, perhaps the formula is intended for n >= 1.Wait, in our earlier derivation, b_n was defined for n >= 0, and a_n for n >= 0. So, perhaps the formula needs to be adjusted for n = 0 and n = 1.Alternatively, maybe the formula is valid for n >= 2, with appropriate definitions for F_{n-1} and F_{n-2}.Given that, let's test n = 2.a_2 = (a_1 a_0 + 1)/(a_1 + a_0) = (y x + 1)/(x + y)Using the formula:F_{n-1} = F_1 = 1F_{n-2} = F_0 = 1 (if we define F_0 = 1)So,a_2 = [ (x + 1)^1 (y + 1)^1 + (x - 1)^1 (y - 1)^1 ] / [ (x + 1)^1 (y + 1)^1 - (x - 1)^1 (y - 1)^1 ]= [ (x + 1)(y + 1) + (x - 1)(y - 1) ] / [ (x + 1)(y + 1) - (x - 1)(y - 1) ]Compute numerator:= (xy + x + y + 1) + (xy - x - y + 1) = 2xy + 2Denominator:= (xy + x + y + 1) - (xy - x - y + 1) = 2x + 2yTherefore,a_2 = (2xy + 2)/(2x + 2y) = (xy + 1)/(x + y)Which matches the given recursive formula. So, the formula works for n = 2.Similarly, for n = 3:a_3 = (a_2 a_1 + 1)/(a_2 + a_1)Using the formula:F_{n-1} = F_2 = 2F_{n-2} = F_1 = 1So,a_3 = [ (x + 1)^1 (y + 1)^2 + (x - 1)^1 (y - 1)^2 ] / [ (x + 1)^1 (y + 1)^2 - (x - 1)^1 (y - 1)^2 ]Compute numerator:= (x + 1)(y^2 + 2y + 1) + (x - 1)(y^2 - 2y + 1)= (x y^2 + 2x y + x + y^2 + 2y + 1) + (x y^2 - 2x y + x - y^2 + 2y - 1)Simplify:= x y^2 + 2x y + x + y^2 + 2y + 1 + x y^2 - 2x y + x - y^2 + 2y - 1Combine like terms:= 2x y^2 + 0x y + 2x + 0y^2 + 4y + 0= 2x y^2 + 2x + 4yDenominator:= (x + 1)(y^2 + 2y + 1) - (x - 1)(y^2 - 2y + 1)= (x y^2 + 2x y + x + y^2 + 2y + 1) - (x y^2 - 2x y + x - y^2 + 2y - 1)= x y^2 + 2x y + x + y^2 + 2y + 1 - x y^2 + 2x y - x + y^2 - 2y + 1Simplify:= 0x y^2 + 4x y + 0x + 2y^2 + 0y + 2= 4x y + 2y^2 + 2Therefore,a_3 = (2x y^2 + 2x + 4y) / (4x y + 2y^2 + 2)Factor numerator and denominator:Numerator: 2(x y^2 + x + 2y)Denominator: 2(2x y + y^2 + 1)Cancel 2:= (x y^2 + x + 2y) / (2x y + y^2 + 1)Now, compute a_3 using the recursive formula:a_3 = (a_2 a_1 + 1)/(a_2 + a_1)We have a_2 = (x y + 1)/(x + y)So,a_3 = [ ( (x y + 1)/(x + y) ) * y + 1 ] / [ (x y + 1)/(x + y) + y ]Simplify numerator:= [ (x y^2 + y)/(x + y) + 1 ] = [ (x y^2 + y + x + y) / (x + y) ] = (x y^2 + x + 2y) / (x + y)Denominator:= [ (x y + 1 + y(x + y)) / (x + y) ] = [x y + 1 + x y + y^2] / (x + y) = (2x y + y^2 + 1) / (x + y)Therefore,a_3 = [ (x y^2 + x + 2y)/(x + y) ] / [ (2x y + y^2 + 1)/(x + y) ] = (x y^2 + x + 2y) / (2x y + y^2 + 1)Which matches the result from the general formula. So, the formula works for n = 3 as well.Therefore, the general term a_n is given by:a_n = frac{(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} + (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}}{(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} - (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}}Where F_n is the nth Fibonacci number with F_0 = 1, F_1 = 1, F_2 = 2, etc.This completes the solution.

💡Okay, so I have this problem about a renovation project that needs various materials, and I need to figure out the total number of truck-loads required. Let me see... The materials listed are sand, dirt, cement, gravel, and crushed stone, each with their own quantities in truck-loads. First, I need to understand what exactly is being asked. It seems straightforward: add up all the truck-loads of each material to get the total. But let me make sure I'm interpreting this correctly. Each material has a certain amount needed, measured in truck-loads, and I need to sum them all up to find the total number of truck-loads required for the project.Alright, so the quantities are:- Sand: 0.17 truck-loads- Dirt: 0.33 truck-loads- Cement: 0.17 truck-loads- Gravel: 0.25 truck-loads- Crushed Stone: 0.08 truck-loadsI think the best way to approach this is to add these numbers one by one. Maybe I'll write them down and add them step by step to minimize errors.Let me start by adding the first two: sand and dirt.0.17 (sand) + 0.33 (dirt) = ?Hmm, 0.17 plus 0.33. Let me think. 0.1 plus 0.3 is 0.4, and 0.07 plus 0.03 is 0.10. So, 0.4 plus 0.10 is 0.50. Okay, so together, sand and dirt require 0.50 truck-loads.Next, I'll add cement to this total. So, 0.50 (sand + dirt) + 0.17 (cement) = ?0.50 plus 0.17. Well, 0.5 plus 0.1 is 0.6, and 0.0 plus 0.07 is 0.07. So, 0.6 plus 0.07 is 0.67. Now, the total is 0.67 truck-loads.Moving on, I need to add gravel. So, 0.67 (previous total) + 0.25 (gravel) = ?0.67 plus 0.25. Let's break it down: 0.6 plus 0.2 is 0.8, and 0.07 plus 0.05 is 0.12. So, 0.8 plus 0.12 is 0.92. Now, the total is 0.92 truck-loads.Finally, I need to add crushed stone. So, 0.92 (previous total) + 0.08 (crushed stone) = ?0.92 plus 0.08. That's straightforward: 0.92 plus 0.08 equals 1.00.Wait a minute, that adds up perfectly to 1.00 truck-loads. That seems neat, but let me double-check to make sure I didn't make any mistakes in my addition.Starting over:0.17 (sand) + 0.33 (dirt) = 0.500.50 + 0.17 (cement) = 0.670.67 + 0.25 (gravel) = 0.920.92 + 0.08 (crushed stone) = 1.00Yes, that seems correct. All the decimals add up properly, and there's no leftover amount. So, the total number of truck-loads needed for the renovation project is exactly 1.00.I wonder if there's a quicker way to verify this. Maybe by rounding each number to the nearest tenth and adding them up to see if it still makes sense.Rounding:- 0.17 ≈ 0.2- 0.33 ≈ 0.3- 0.17 ≈ 0.2- 0.25 ≈ 0.3- 0.08 ≈ 0.1Adding these rounded numbers: 0.2 + 0.3 + 0.2 + 0.3 + 0.1 = 1.1Hmm, that's 1.1, which is slightly higher than the exact total of 1.00. But considering we rounded up some numbers, it makes sense that the estimate is a bit higher. It still gives me confidence that 1.00 is the correct exact total.Another way to think about it is to consider the place values. Let's line them up by decimal points and add:``` 0.17+0.33+0.17+0.25+0.08-------```Adding the hundredths place: 7 + 3 + 7 + 5 + 8 = 30. So, that's 0.30.Adding the tenths place: 1 + 3 + 1 + 2 + 0 = 7. So, that's 0.70.Adding the units place: All are 0, so nothing there.Now, adding 0.70 (tenths) + 0.30 (hundredths) = 1.00.Yes, that confirms it again. The total is exactly 1.00 truck-loads.I think I've verified it enough ways. It seems consistent across different methods of addition and rounding checks. So, I'm pretty confident that the total number of truck-loads needed is 1.00.

💡Okay, so I have this problem about a convex quadrilateral ABCD. The given information is that the length of diagonal BD is 2, and the dot product of vectors AC and BD is zero. Also, there's this other equation involving the dot product of two vector sums: (AB + DC) · (BC + AD) equals 5. I need to find the area of quadrilateral ABCD.First, let me try to visualize the quadrilateral. Since it's convex, all its interior angles are less than 180 degrees, and the diagonals AC and BD intersect each other inside the quadrilateral. The fact that BD is 2 units long is straightforward. The dot product of AC and BD being zero tells me that these two diagonals are perpendicular to each other. So, AC ⊥ BD.Now, the area of a convex quadrilateral with perpendicular diagonals can be found using the formula: Area = (1/2) * |AC| * |BD|. Since BD is given as 2, if I can find the length of AC, I can plug it into this formula and find the area.So, my goal now is to find the length of AC. The other given equation is (AB + DC) · (BC + AD) = 5. Hmm, this looks a bit complicated, but maybe I can simplify it by expressing the vectors in terms of the diagonals AC and BD.Let me recall that in a quadrilateral, the sum of vectors around the perimeter can sometimes be expressed in terms of the diagonals. Let me write down each vector in terms of position vectors. Let me assign position vectors to the points: Let’s say A is at vector **a**, B at **b**, C at **c**, and D at **d**.So, vector AB is **b** - **a**, vector DC is **c** - **d**, vector BC is **c** - **b**, and vector AD is **d** - **a**.So, let's compute (AB + DC) and (BC + AD):AB + DC = (**b** - **a**) + (**c** - **d**) = (**b** + **c**) - (**a** + **d**)BC + AD = (**c** - **b**) + (**d** - **a**) = (**c** + **d**) - (**a** + **b**)So, the dot product is [(**b** + **c**) - (**a** + **d**)] · [(**c** + **d**) - (**a** + **b**)] = 5.Let me denote **s** = **a** + **b** + **c** + **d**. Hmm, not sure if that helps, but let's see.Alternatively, maybe I can express these in terms of the diagonals AC and BD.Wait, AC is **c** - **a**, and BD is **d** - **b**. Since AC · BD = 0, we know they are perpendicular.Let me see if I can express the vectors AB + DC and BC + AD in terms of AC and BD.Let me compute AB + DC:AB = **b** - **a**DC = **c** - **d**So, AB + DC = (**b** - **a**) + (**c** - **d**) = (**b** + **c**) - (**a** + **d**)Similarly, BC + AD:BC = **c** - **b**AD = **d** - **a**So, BC + AD = (**c** - **b**) + (**d** - **a**) = (**c** + **d**) - (**a** + **b**)Hmm, so both these sums involve combinations of position vectors. Maybe I can relate them to the diagonals.Wait, the diagonals are AC = **c** - **a** and BD = **d** - **b**. Let me see if I can express (**b** + **c**) - (**a** + **d**) in terms of AC and BD.Let me compute (**b** + **c**) - (**a** + **d**) = (**b** - **a**) + (**c** - **d**) = AB + DC, which is the same as before. Maybe I need a different approach.Alternatively, let me consider that (**b** + **c**) - (**a** + **d**) can be written as ( (**b** - **a**) + (**c** - **d**) ) = AB + DC.Similarly, (**c** + **d**) - (**a** + **b**) = ( (**c** - **a**) + (**d** - **b**) ) = AC + BD.Wait, that seems useful. So, BC + AD = AC + BD.So, putting it all together, the dot product becomes:(AB + DC) · (BC + AD) = (AB + DC) · (AC + BD) = 5.But AB + DC is equal to (**b** + **c**) - (**a** + **d**). Let me see if I can express this in terms of AC and BD as well.Wait, (**b** + **c**) - (**a** + **d**) = ( (**b** - **a**) + (**c** - **d**) ) = AB + DC.But AC = **c** - **a**, and BD = **d** - **b**. So, BD = **d** - **b**, which implies **d** = **b** + BD.Similarly, AC = **c** - **a**, so **c** = **a** + AC.Let me substitute these into AB + DC:AB = **b** - **a**DC = **c** - **d** = (**a** + AC) - (**b** + BD) = (**a** - **b**) + AC - BD.So, AB + DC = (**b** - **a**) + ( (**a** - **b**) + AC - BD ) = (**b** - **a** + **a** - **b**) + AC - BD = 0 + AC - BD = AC - BD.So, AB + DC = AC - BD.Similarly, earlier, we saw that BC + AD = AC + BD.Therefore, the dot product (AB + DC) · (BC + AD) becomes (AC - BD) · (AC + BD) = 5.This simplifies to (AC · AC) + (AC · BD) - (BD · AC) - (BD · BD) = |AC|² - |BD|², since AC · BD = BD · AC, and they are equal, so they subtract out.But wait, we know that AC · BD = 0, as given in the problem. So, actually, the cross terms are zero.Therefore, (AC - BD) · (AC + BD) = |AC|² - |BD|² = 5.Given that BD = 2, so |BD|² = 4. Therefore, |AC|² - 4 = 5, so |AC|² = 9, which means |AC| = 3.Great, so now we know that AC is 3 units long, and BD is 2 units long, and they are perpendicular to each other.Therefore, the area of quadrilateral ABCD is (1/2) * |AC| * |BD| = (1/2) * 3 * 2 = 3.So, the area is 3.**Final Answer**The area of quadrilateral (ABCD) is boxed{3}.

💡Alright, so I've got this problem here where I need to find a number in a 5x5 array that is the largest in its column and also the smallest in its row. The array is:20 5 9 4 3 22 12 25 19 15 13 6 9 10 18 23 11 30 21 7 18 8 10 14 6 And the options are A) 7, B) 12, C) 18, D) 23, E) 30.Okay, so first, I need to understand what the problem is asking. It wants a number that is the largest in its column, meaning if I look at the column it's in, it's bigger than all the other numbers in that column. At the same time, it should be the smallest in its row, meaning when I look at the row it's in, it's smaller than all the other numbers in that row.So, I think the best way to approach this is to first identify the largest numbers in each column and then check if any of those are also the smallest in their respective rows.Let me start by labeling the rows and columns for clarity. I'll number the rows from top to bottom as Row 1 to Row 5, and the columns from left to right as Column 1 to Column 5.So, the array looks like this with row and column labels:Row 1: 20 (C1), 5 (C2), 9 (C3), 4 (C4), 3 (C5) Row 2: 22 (C1), 12 (C2), 25 (C3), 19 (C4), 15 (C5) Row 3: 13 (C1), 6 (C2), 9 (C3), 10 (C4), 18 (C5) Row 4: 23 (C1), 11 (C2), 30 (C3), 21 (C4), 7 (C5) Row 5: 18 (C1), 8 (C2), 10 (C3), 14 (C4), 6 (C5)Now, I'll go through each column and find the largest number in that column.Starting with Column 1: 20, 22, 13, 23, 18. The largest here is 23.Column 2: 5, 12, 6, 11, 8. The largest is 12.Column 3: 9, 25, 9, 30, 10. The largest is 30.Column 4: 4, 19, 10, 21, 14. The largest is 21.Column 5: 3, 15, 18, 7, 6. The largest is 18.So, the largest numbers in each column are: 23 (C1), 12 (C2), 30 (C3), 21 (C4), 18 (C5).Now, I need to check if any of these numbers are also the smallest in their respective rows.Starting with 23 in Column 1, which is in Row 4. Looking at Row 4: 23, 11, 30, 21, 7. The smallest number here is 7, not 23. So, 23 is not the smallest in its row.Next, 12 in Column 2, which is in Row 2. Looking at Row 2: 22, 12, 25, 19, 15. The smallest number here is 12. So, 12 is the smallest in its row and the largest in its column. That seems to fit the criteria.Just to be thorough, I'll check the others too.30 in Column 3 is in Row 4. Row 4: 23, 11, 30, 21, 7. The smallest is 7, not 30.21 in Column 4 is in Row 4. Again, the smallest in Row 4 is 7, not 21.18 in Column 5 is in Row 3. Row 3: 13, 6, 9, 10, 18. The smallest is 6, not 18.So, the only number that is both the largest in its column and the smallest in its row is 12.Therefore, the answer should be B) 12.

💡Okay, so I have this probability problem here, and I need to figure out the chance that Abe and Bob show the same color jelly bean. Let me try to break this down step by step.First, let me list out what each person has:- Abe has 1 green and 2 red jelly beans. So, in total, Abe has 3 jelly beans.- Bob has 2 green, 1 yellow, and 1 red jelly bean. That makes 4 jelly beans in total for Bob.- Cal has 1 green, 2 yellow, and 2 red jelly beans, totaling 5 jelly beans. But wait, the problem is only asking about Abe and Bob, so maybe I don't need to worry about Cal? Hmm, the question says each of them randomly picks a jelly bean to show the others, but it specifically asks about Abe and Bob matching. So, maybe Cal's jelly beans don't affect the probability between Abe and Bob. I think I can ignore Cal for this problem.So, focusing on Abe and Bob:Abe has 1 green and 2 red. So, the probability that Abe picks green is 1 out of 3, and red is 2 out of 3.Bob has 2 green, 1 yellow, and 1 red. So, the probability that Bob picks green is 2 out of 4, which simplifies to 1 out of 2. The probability Bob picks red is 1 out of 4.Now, I need to find the probability that both Abe and Bob pick the same color. That can happen in two ways: both pick green or both pick red.Let me calculate each scenario:1. Both pick green: - Probability Abe picks green: 1/3 - Probability Bob picks green: 1/2 - So, the combined probability is (1/3) * (1/2) = 1/62. Both pick red: - Probability Abe picks red: 2/3 - Probability Bob picks red: 1/4 - So, the combined probability is (2/3) * (1/4) = 2/12 = 1/6Now, to get the total probability that Abe and Bob show the same color, I add these two probabilities together:1/6 (both green) + 1/6 (both red) = 2/6 = 1/3So, the probability that Abe and Bob show the same color is 1/3.Wait, let me double-check that. Did I consider all possible cases? I considered both green and both red. Is there any other color they could match on? Well, Abe doesn't have any yellow, and Bob doesn't have any other colors besides green, yellow, and red. So, the only possible matches are green and red.Another way to think about it is to list all possible outcomes. Abe has 3 choices, and Bob has 4 choices, so there are 3 * 4 = 12 possible pairs of jelly beans they could show.Let me list them:Abe's choices: G, R, RBob's choices: G, G, Y, RSo, the possible pairs are:1. G (Abe) and G (Bob)2. G (Abe) and G (Bob)3. G (Abe) and Y (Bob)4. G (Abe) and R (Bob)5. R (Abe) and G (Bob)6. R (Abe) and G (Bob)7. R (Abe) and Y (Bob)8. R (Abe) and R (Bob)9. R (Abe) and G (Bob)10. R (Abe) and G (Bob)11. R (Abe) and Y (Bob)12. R (Abe) and R (Bob)Now, let's count the number of matching colors:- Both green: There are two Gs for Bob, so when Abe picks G, Bob can pick G in two ways. So, that's 2 matches.- Both red: Abe has two Rs, and Bob has one R. So, when Abe picks R, Bob can pick R in one way. Since Abe has two Rs, that's 2 * 1 = 2 matches.Wait, but in the list above, how many times do we have both green or both red?Looking at the list:1. G & G - match2. G & G - match3. G & Y - no4. G & R - no5. R & G - no6. R & G - no7. R & Y - no8. R & R - match9. R & G - no10. R & G - no11. R & Y - no12. R & R - matchSo, matches are at positions 1, 2, 8, and 12. That's 4 matches out of 12 possible pairs.So, 4/12 simplifies to 1/3.Okay, that confirms my earlier calculation. So, the probability is indeed 1/3.I think that's solid. I considered both the probability approach and the enumeration of possible outcomes, and both lead to the same result. So, I'm confident that the answer is 1/3.**Final Answer**boxed{C}

💡First, I need to determine the total number of parts in the original ratio. The ratio consists of 7 parts blue paint, 2 parts red paint, and 1 part white paint, which adds up to 10 parts in total.Next, Marla adds 140 ounces of blue paint, which corresponds to the 7 parts of blue paint in the ratio. To find out how much one part is in ounces, I divide the amount of blue paint by the number of blue parts: 140 ounces divided by 7 parts equals 20 ounces per part.Now, considering the total mixture should not exceed 180 ounces, I calculate the maximum number of parts that can be used in the mixture by dividing the total mixture limit by the ounces per part: 180 ounces divided by 20 ounces per part equals 9 parts total.Since the white paint corresponds to 1 part of the original ratio, and we have a maximum of 9 parts for the total mixture, I calculate the amount of white paint by multiplying the number of white parts by the ounces per part: 1 part white paint multiplied by 20 ounces per part equals 20 ounces of white paint.Therefore, Marla should add 20 ounces of white paint to the mixture.