数列an＝n平方，求Sn〔1⼀6n（n＋1）（2n＋1）〕怎么推导，顺便讲下三次方的推导

an = n^2
= n(n+1)- n
= (1/3)[ n(n+1)(n+2) - (n-1)n(n+1) ] - (1/2) [ n(n+1) - (n-1)n]
Sn =a1+a2+...+an
=(1/3)n(n+1)(n+2) - (1/2)n(n+1)
= (1/6)n(n+1)(2n+1)
bn =n^3
= n(n+1)(n+2) - 3n^2-2n
=n(n+1)(n+2) - 3n(n+1) +n
= (1/4)[ n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)] - [n(n+1)(n+2) -(n-1)n(n+1)]
+ (1/2)[ n(n+1) -(n-1)n ]
Tn = b1+b2+...+bn
=(1/4)n(n+1)(n+2)(n+3) -n(n+1)(n+2) + (1/2)n(n+1)
= (1/4)n(n+1)[ (n+2)(n+3) - 4(n+2) + 2]
= (1/4)n(n+1)( n^2+n)
= [(1/2)n(n+1)]^2