本帖最后由 tommywong 于 2021-12-6 08:10 编辑
$a_{k+1}=a$
$\displaystyle \sum_{i=1}^k \frac{1}{a-a_i}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
=-\sum_{i=1}^k \prod_{j=1\atop j\neq i}^{k+1} \frac{1}{a_i-a_j}$
$\displaystyle =\prod_{j=1}^k \frac{1}{a-a_j}-\sum_{i=1}^{k+1} \prod_{j=1\atop j\neq i}^{k+1} \frac{1}{a_i-a_j}=\prod_{j=1}^k \frac{1}{a-a_j}$
$\displaystyle \sum_{i=1}^k \frac{1}{(a-a_i)^2}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
=\left(\sum_{j=1}^k \frac{1}{a-a_j}\right)
\left(\prod_{j=1}^k \frac{1}{a-a_j}\right) ?$
$\displaystyle \sum_{i=1}^k \frac{1}{(a-a_i)^3}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
= ?$ |
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发表于 2021-12-5 22:47
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发表于 2021-12-6 01:28
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