每日一题：2020-07-29

发表于 更新于 分类于

每日一题: 2020-07-29

题目: 已知可把求和a1+a2+⋯+ana_1+a_2+\cdots +a_n简写为∑i=1nai\sum\limits_{i=1}^na_i, 即a1+a2+⋯an=∑i=1naia_1+a_2+\cdots a_n=\sum\limits_{i=1}^n a_i.
假设Sk=∑i=1kai;(k=1,2,…,n)S_k=\sum\limits_{i=1}^ka_i; (k=1,2,\ldots,n) 请证明:

∑k=1nakbk=Snbn+∑k=1n−1Sk(bk−bk+1)\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S0=0S_0=0, 则a1=S1−S0,ak=Sk−Sk−1(k=2,3,…,n)a_1=S_1-S_0, a_k=S_k-S_{k-1} (k=2,3,\ldots,n)

∴∑k=1nakbk=∑k=1nbk(Sk−Sk−1)=∑k=1nbkSk−∑k=1nbkSk−1\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

∑k=1nbkSk−1=∑k=2nbkSk−1=∑k=1n−1bk+1Sk\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

∑k=1nakbk=Snbn+∑k=1n−1bkSk−∑k=1n−1bk+1Sk=Snbn+∑k=1n−1Sk(bk−bk+1)\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})