数列“an=n平方”的前n项和为什么是Sn=n(n+1)(2n+1)/6呢
问题描述：
数列“an=n平方”的前n项和为什么是Sn=n(n+1)(2n+1)/6呢
答
a(n)=n^2 = n(n+1) - n ,
n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3,
n = [n(n+1) - (n-1)n]/2,
a(n) = [n(n+1)(n+2) - (n-1)n(n+1)]/3 - [ n(n+1) - (n-1)n]/2,
s(n) = [1*2*3-0 + 2*3*4-1*2*3 + ... + (n-1)n(n+1)-(n-2)(n-1)n + n(n+1)(n+2)-(n-1)n(n+1)]/3 -
- [1*2-0 + 2*3-1*2 +... + (n-1)n-(n-2)(n-1) + n(n+1)-(n-1)n]/2
=n(n+1)(n+2)/3 - n(n+1)/2
=n(n+1)/6[2n+4-3]
=n(n+1)(2n+1)/6
答
历代数学家辛苦推导出来的公式,不要怀疑吧?不妨用数学归纳法自行证明一下.