Question

1的2次方+2的2次方+3的2次方+…+N的N次方

Answer 1

n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n
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
各等式全相加
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+3^2+...+n^2=n(n+1)(2n+1)/6

Answer 2

1²+2²+3²+……+n²
(n+1)³-n³=3n²+3n+1
n³-(n-1)³=3(n-1)²+3(n-1)+1
...
..
...
2³-1³=3*1²+3*1+1
把上面n个式子相加得：(n+1)³-1
=
3*
[1²+2²+...+n²]
+3*[1+2+....+n]
+n
(1/3)*[(n+1)³-1-n-(1/2)*n(n+1)]=(1/6)n(n+1)(2n+1)