吉米多维奇数学分析中一道习题的分享
证明:12+22+⋯+n2=n(n+1)(2n+1)61^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}12+22+⋯+n2=6n(n+1)(2n+1)
两种方法：
- 数学归纳法
证明: 当n=1n=1n=1时,等式显然成立
设n=kn=kn=k时,等式成立,即12+22+⋯+k2=k(k+1)(2k+1)6+(k+1)21^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^212+22+⋯+k2=6k(k+1)(2k+1)+(k+1)2,则对于n=k+1n=k+1n=k+1时,有
12+22+⋯+k2+(k+1)2=k(k+1)(2k+1)6+(k+1)2=16(k+1)[k(2k+1)+6(k+1)]=(k+1)[(k+1)+1][2(k+1)+1]6 \begin{aligned} 1^2+2^2+\cdots+k^2+(k+1)^2&=\frac{k(k+1)(2k+1)}{6}+(k+1)^2 \\ &=\frac{1}{6}(k+1)[k(2k+1)+6(k+1)]=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6} \end{aligned} 12+22+⋯+k2+(k+1)2=6k(k+1)(2k+1)+(k+1)2=61(k+1)[k(2k+1)+6(k+1)]=6(k+1)[(k+1)+1][2(k+1)+1]
即对于n=k+1n=k+1n=k+1时,等式也成立,于是,对于任何正整数nnn,有12+22+⋯+n2=n(n+1)(2n+1)61^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}12+22+⋯+n2=6n(n+1)(2n+1)
2.直接验证法
首先(n+1)3=n3+3n2+3n+1(n+1)^3=n^3+3n^2+3n+1(n+1)3=n3+3n2+3n+1即n2=13[(n+1)3−n3−3n−1]n^2=\frac{1}{3}[(n+1)^3-n^3-3n-1]n2=31[(n+1)3−n3−3n−1]
那么求和可以写为:
13[23−13−3−1+33−23−3∗2−1+⋯+(n+1)3−n3−3n−1] \frac{1}{3}\left[ 2^3-1^3-3-1+3^3-2^3-3*2-1+\cdots +(n+1)^3-n^3-3n-1 \right] 31[23−13−3−1+33−23−3∗2−1+⋯+(n+1)3−n3−3n−1]
显然三次方项都被相互抵消，−3−3∗2+3∗3−⋯-3-3*2+3*3-\cdots−3−3∗2+3∗3−⋯是等差数列，−1−1−1−1−⋯-1-1-1-1-\cdots−1−1−1−1−⋯是常数列.那么上式改写为:
=13[(n+1)3+13−n−3n(n+1)2]=133n2+3n−2n−22（通分）=2n3+3n2+n6=n(n+1)(2n+1)6 \begin{aligned} &=\frac{1}{3}\left[ (n+1)^3+1^3-n-\frac{3n(n+1)}{2}\right] \\ &=\frac{1}{3}\frac{3n^2+3n-2n-2}{2} （通分） \\ &=\frac{2n^3+3n^2+n}{6} \\ &=\frac{n(n+1)(2n+1)}{6}\\ \end{aligned} =31[(n+1)3+13−n−23n(n+1)]=3123n2+3n−2n−2（通分）=62n3+3n2+n=6n(n+1)(2n+1)
上式得证
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