已知 a(n) 与 a(n+1) 是方程 x^2-b(n)x+1/2^n=0 的两根，a(1)=1 ，求 ∑(k=1,∞)b(k)

wintex

請問代數

elim

题：设\(\;a_1{\small=1,}\,a_na_{n+1}=\frac{1}{2^n},\, b_n=a_n+a_{n+1}\,\small(n\in\mathbb{N}^+).\,\)求\(\,\sum b_n\)
解：\(\because\;\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n+1}{2}\rfloor=\lfloor\frac{n}{2}+\frac{n+1}{2}\rfloor=n\;(\forall n\in\mathbb{Z}),\quad a_1=1,\)
\(\therefore\quad a_n=2^{-\lfloor\frac{n}{2}\rfloor},\;\;{\displaystyle{\small\sum_{n=1}^{\infty}}b_n=1+{\small2\sum_{k=1}^{\infty}}}{(a_{2k}+a_{2k+1})}=5.\quad\small\square\)

elim

注记：\(\displaystyle\,\sum_{n=1}^{\infty}b_n = (a_1+a_2)+(a_2+a_3)+(a_3+a_4)+\cdots\)
\(\qquad=a_1+2(a_2+a_3)+2(a_4+a_5)+\cdots+2(a_{2k}+a_{2k+1})+\cdots\)
\(\qquad\displaystyle=a_1+2{\small\sum_{k=1}^{\infty}}(a_{2k}+a_{2k+1})=a_1+4{\small\sum_{k=1}^{\infty}\frac{1}{2^k}}=5\)

luyuanhong

楼上 elim 的解答很好！已收藏。下面是此题的详细解答过程：

elim

题：设\(\;a_1{\small=1,}\,a_na_{n+1}=\frac{1}{2^n},\, b_n=a_n+a_{n+1}\,\small(n\in\mathbb{N}^+).\,\)求\(\,\sum b_n\)
解：\(\because\;\lfloor x+m\rfloor = \lfloor x\rfloor+m\;{\small(\forall m\in\mathbb{Z})},\quad a_1=1=2^{-\lfloor\frac{1}{2}\rfloor},\)
\(\qquad\therefore\;2^{-\lfloor\frac{n}{2}\rfloor}2^{-\lfloor\frac{n+1}{2}\rfloor}=2^{-\lfloor\frac{n}{2}+\frac{n+1}{2}\rfloor}=2^{-\lfloor n+\frac{1}{2}\rfloor}={\large\frac{1}{2^n}}\;\small(\forall n\in\mathbb{N}).\)
\(\qquad\boxed{a_n=2^{-\lfloor\frac{n}{2}\rfloor},\;a_{2n}=a_{2n+1}=\small\frac{\large_1}{2^n}.}\)
\(\qquad\displaystyle{\small\sum_{n=1}^{\infty}}b_n\small=(a_1+a_2)+(a_2+a_3)+(a_3+a_4)+\cdots \)
\(\qquad\qquad\;\small\,= a_1+2(a_2+a_3)+2(a_4+a_5)+\cdots+2(a_{2k}+a_{2k+1})+\cdots\)
\(\qquad\qquad\;\displaystyle{\small=a_1+2\sum_{k=1}^{\infty}}{(a_{2k}+a_{2k+1})}=a_1{\small+\,4\sum_{k=1}^{\infty}}a_{2k}=5.\quad\small\square\)