Question

高中数学 数列求和

如图

Answer 1

先知道基础公式:
通项为 n^2 数列的前n项和为 1/6*n*(n+1)*(2n+1)
通项为 n 数列的前n项和为 1/2*n*(n+1)

所以通项为 2*n^2-2n+1的前n项和为:
2*1/6*n*(n+1)*(2n+1)-2*1/2*n*(n+1)+n
即为:
1/3*n*(n+1)*(2n+1)-n*(n+1)+n

Answer 2

没图

Answer 3

公式：
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
1+2+3+...+n=n(n+1)/2

Sn=a1+a2+...+an
=2(1^2+2^2+...+n^2)-2(1+2+...+n)+n
=2n(n+1)(2n+1)/6-2n(n+1)/2+n
=n(n+1)[(2n+1)/3-1]+n
=n(n+1)(2n-2)/3+n
=(n/3)(2n^2-2+3)
=(n/3)(2n^2+1)
=n(2n^2+1)/3

Sn=n(2n^2+1)/3

Answer 4

an = 2n(n-1)+1 = 2/3((n+1)-(n-2))n(n-1)+1
= 2/3((n+1)n(n-1) - n(n-1)(n-2)) + 1

所以

a1 = 2/3(0 - 0) + 1
a2 = 2/3(6 - 0) + 1
a3 = 2/3(24 - 6) + 1
....
an = 2/3((n+1)n(n-1) - n(n-1)(n-2)) + 1

相加得：
Sn = 2/3((n+1)n(n-1)-0) + n
= 1/3 * n(2n*n - 2 +3)
= n(2n^2 + 1)/3