多阶等差数列前n项和的推导
多阶等差数列前n项和的推导
其中一阶就是等差，二阶每一项的差成等差，三阶是每一项差的差成等差以此类推，一阶形如An=n,二阶形如An=n^2等等。就推一下二阶的吧
法一：
a(n)=n^2 = n(n+1) - n = [n(n+1)(n+2)-(n-1)n(n+1)]/3 - [n(n+1)-(n-1)n]/2,
s(n) = a(1)+a(2)+...+a(n-1)+a(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)(2n+4-3)/6
= n(n+1)(2n+1)/6
法二：
s(n) = 1^2 + 2^2 + ...+ n^2 .
(n+1)^3 = n^3 + 3n^2 + 3n + 1,
(n+1)^3 - n^3 = 3n^2 + 3n + 1.
n^3 - (n-1)^3 = 3(n-1)^2 + 3(n-1) + 1,
(n-1)^3 - (n-2)^3 = 3(n-2)^2 + 3(n-2) + 1,
...
3^3 - 2^3 = 3*2^2 + 3*2 + 1,
2^3 - 1^3 = 3*1^2 + 3*1 + 1,
(n+1)^3 - 1^3 = 3[1^2 + 2^2 + ...+ n^2] + 3[1+2+...+n] + n
=3s(n) + 3n(n+1)/2 + n
3s(n) = (n+1)^3 - 1 - 3n(n+1)/2 - n = n^3 + 3n^2 + 2n - 3n(n+1)/2
= n^2(n+1) + 2n(n+1) - 3n(n+1)/2
= n(n+1)[2n + 4 - 3]/2
= n(n+1)(2n+1)/2,
s(n) = n(n+1)(2n+1)/6.