- \(a_n = n^2\)
首先从一个等式入手:
\[ (n+1)^3 - n^3 = 3n^2 + 3n + 1 \]
那么直接求和显然也成立了:
\[ \sum_{i=1}^n (i+1)^3 - \sum_{i=1}^n i^3 = 3\sum_{i=1}^n i^2 + 3 \sum_{i=1}^n i + \sum_{i=1}^n\]
左边直接错位相减,右边第一项为所求,设为\(S_n\),第二项等差数列求和\(\frac{n(n+1)}{2}\),第三项就是\(n\).
\[ (n+1)^3 - 1 = 3S_n + \frac{3n(n+1)}{2} + n \]
\[S_n = \frac{n(n+1)(2n+1)}{6}\]
2.\(a_n = n^3\)
和上一个差不多:
\[ (n+1)^4 - n^4 = 4n^3 +6n^2 + 4n + 1\]
然后几乎完全一致的操作一下:
\[ \sum_{i=1}^n (i+1)^4 - \sum_{i=1}^n i^4 = 4\sum_{i=1}^n i^3 + 6\sum_{i=1}^n i^2 + 4\sum_{i=1}^n i + \sum_{i=1}^n\]
\[ S_n = (\frac{n(n+1)}{2})^2 \]
\(To \ \ be \ \ continued\)