Question

从1的平方一直加到100的平方是多少

Answer 1

1^2=1*2-1
2^2=2*3-2
.....
.....
n^2=n(n+1)-n
由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+...+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+....+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3

所以1^2+2^2+3^2+......+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]

或者数学归纳法..或者
2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
......
n^3-(n-1)^3=2*n^2+(n-1)^2-n
等式全相加
n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)

n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)

n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1

n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2

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

(1^2+2^2+...+n^2)=n(n+1)(2n+1)/6
当n=100时(1^2+2^2+...+100^2)=100(100+1)(2*100+1)/6
=100*101*201/6
=50*101*67
=338350

Answer 2

1＾2＋2＾2＋．．．．＋100^2=
338350
1＾2＋2＾2＋．．．．＋n^2=
1/6 n (1 + n) (1 + 2 n)