每日一题：2020-07-29

每日一题: 2020-07-29

题目: 已知可把求和$a1+a_2+\cdots +a_n$简写为$\sum\limits{i=1}^nai$, 即$a_1+a_2+\cdots a_n=\sum\limits{i=1}^n ai$.
假设$S_k=\sum\limits{i=1}^ka_i; (k=1,2,\ldots,n)$ 请证明:

$$\sum_{k=1}^na_kb_k=S_nb_n+\sum_{k=1}^{n-1}S_k(b_k-b_{k+1})$$

参考思路

令$S_0=0$, 则$a_1=S_1-S_0, a_k=S_k-S_{k-1} (k=2,3,\ldots,n)$ $$ \therefore \sum_{k=1}^na_kb_k=\sum_{k=1}^n b_k(S_k-S_{k-1})=\sum_{k=1}^nb_kS_k-\sum_{k=1}^nb_kS_{k-1} $$ 又 $$ \sum_{k=1}^nb_kS_{k-1}=\sum_{k=2}^n b_kS_{k-1}=\sum_{k=1}^{n-1}b_{k+1}S_k $$ 所以有 $$ \sum_{k=1}^n a_kb_k=S_nb_n+\sum_{k=1}^{n-1}b_kS_k-\sum_{k=1}^{n-1}b_{k+1}S_k=S_nb_n+\sum_{k=1}^{n-1}S_k(b_k-b_{k+1}) $$