Fancy limit
Author: anonymous
Problem has been solved: 31 times
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The sequence of Fibonacci numbers ${F_n}$ is defined in the following way:
\[F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_{n} \ (n = 1, 2, \ldots).\]
For any positive integer $n$, let's denote by $x_n$ the sum
\[\sum_{k=1}^{n} \frac{3F_{k+1}^2 (F_k^2 + F_{k+1}^2 - F_{k+3}^2) - 2F_k^2F_{k+3}^2 - F_{k+2}^2(F_{k+2}^2 + F_{k+3}^2 - F_k^2)}{F_{2k+1} F_{2k+3} F_{2k+5}}.\]
Let $\dfrac{a}{b} = \displaystyle (\lim_{n\to\infty} x_n)^2$ where $a, b$ are coprime positive integers. Find $a + b$.
Последовательность чисел Фибоначчи ${F_n}$ определена следующим образом:
\[ F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_{n} \ (n = 1, 2, \ldots).\]
Для любого натурального $n$, через $x_n$ обозначим сумму
\[\sum_{k=1}^{n} \frac{3F_{k+1}^2 (F_k^2 + F_{k+1}^2 - F_{k+3}^2) - 2F_k^2F_{k+3}^2 - F_{k+2}^2(F_{k+2}^2 + F_{k+3}^2 - F_k^2)}{F_{2k+1} F_{2k+3} F_{2k+5}}.\]
Пусть $\dfrac{a}{b} = \displaystyle (\lim_{n\to\infty} x_n)^2$, где $a, b$ --- взаимнопростые натуральные числа. Найдите $a + b$.
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