用完全归纳法证明1^2+2^2+...+n^2
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/11 19:18:04
用完全归纳法证明1^2+2^2+...+n^2
![用完全归纳法证明1^2+2^2+...+n^2](/uploads/image/z/19649016-0-6.jpg?t=%E7%94%A8%E5%AE%8C%E5%85%A8%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E1%5E2%2B2%5E2%2B...%2Bn%5E2)
1^2+2^2+...+n^2=n(n+1)(2n+1)/6
n=1,略
假设n=k成立
1^2+2^2+...+k^2=k(k+1)(2k+1)/6
则n=k+1
1^2+2^2+...+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)[k(2k+1)+6(k+1)]/6
=(k+1)[2k^2+7k+6)/6
=(k+1)(k+2)(2k+3)/6
=(k+1)[(k+1)+1][2(k+1)+1]/6
综上
1^2+2^2+...+n^2=n(n+1)(2n+1)/6
n=1,略
假设n=k成立
1^2+2^2+...+k^2=k(k+1)(2k+1)/6
则n=k+1
1^2+2^2+...+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)[k(2k+1)+6(k+1)]/6
=(k+1)[2k^2+7k+6)/6
=(k+1)(k+2)(2k+3)/6
=(k+1)[(k+1)+1][2(k+1)+1]/6
综上
1^2+2^2+...+n^2=n(n+1)(2n+1)/6