Library / / 525 of 1394
, 2016 Difficulty 5.2 AIME, harder Prove it United States
Problem:
An infinite sequence of real numbers a 1 , a 2 , … a_{1}, a_{2}, \ldots a 1 , a 2 , … a n + 3 = a n + 2 − 2 a n + 1 + a n
a_{n+3}=a_{n+2}-2 a_{n+1}+a_{n}
a n + 3 = a n + 2 − 2 a n + 1 + a n n n n a 1 = a 3 = 1 a_{1}=a_{3}=1 a 1 = a 3 = 1 a 98 = a 99 a_{98}=a_{99} a 98 = a 99 a 1 + a 2 + ⋯ + a 100 a_{1}+a_{2}+\cdots+a_{100} a 1 + a 2 + ⋯ + a 100
Proposed by: Evan Chen
Solution Solution:
∑ k = 1 n a k = a 1 + a 2 + a 3 + ∑ k = 1 n − 3 ( a k − 2 a k + 1 + a k + 2 ) = a 1 + a 2 + a 3 + ∑ k = 1 n − 3 a k − 2 ∑ k = 2 n − 2 a k + ∑ k = 3 n − 1 a k = 2 a 1 + a 3 − a n − 2 + a n − 1
\begin{aligned}
\sum_{k=1}^{n} a_{k} & =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3}\left(a_{k}-2 a_{k+1}+a_{k+2}\right) \\
& =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3} a_{k}-2 \sum_{k=2}^{n-2} a_{k}+\sum_{k=3}^{n-1} a_{k} \\
& =2 a_{1}+a_{3}-a_{n-2}+a_{n-1}
\end{aligned}
k = 1 ∑ n a k = a 1 + a 2 + a 3 + k = 1 ∑ n − 3 ( a k − 2 a k + 1 + a k + 2 ) = a 1 + a 2 + a 3 + k = 1 ∑ n − 3 a k − 2 k = 2 ∑ n − 2 a k + k = 3 ∑ n − 1 a k = 2 a 1 + a 3 − a n − 2 + a n − 1
Putting n = 100 n=100 n = 100
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