Question

计算1³+2³+3³+......+99³+100³

Answer 1

公式：1.1³+2³+3³+……+（n-1）³+n³=1/4（n-1）²（n）²；

结果：2. 1³+2³+3³+……+99³+100³=1/4*100²*101²=25502500

推导：设S=1³+2³+3³+4 ³+……..+99³+100³
(n+1)^4=n^4+4n^3+6n^2+4n+1;
n^4=(n-1)^4+4(n-1)^3+6(n-1)^2+4(n-1)+1;
.......
2^4=1^4+4*1^3+6*1^2+4*1+1;
结合自然数逐项平方和公式s2=1/6*n*(n+1)*(2n+1);
与自然数逐项求和公式:s1=1/2*n*(n+1)逐式相加得：
(n+1)^4=1+4S+6*1/6*n*(n+1)*(2n+1)+4*1/2*n*(n+1)+n;
整理既得，S=[1/2*n*(n+1)]^2=s1^2=5050^2=25502500

Answer 2

这有现成公式：1^3+2^3+3^3+.......+n^3=[n(n+1)/2]^2 ，
所以 1^3+2^3+.........+100^3=(100*101/2)^2=5050^2=25502500 。

Answer 3

Sn=n^2(n+1)^2/2
设bn=(n+1)^4-n^4,Bn是bn的前n项和
则Bn=(n+1)^4-1
又
bn=4n^3+6n^2+4n+1
所以Bn=4Sn+6*(1/6)n(n+1)(2n+1)+4*(1/2)n(n+1)+n
(n+1)^4-1=4Sn+n(n+1)(2n+1)+2n(n+1)+n
4Sn=(n+1)^4-[n(n+1)(2n+1)+2n(n+1)+(n+1)]
=(n+1)^4-[n(n+1)(2n+1)+(2n+1)(n+1)]=(n+1)^4-[(n+1)(2n+1)(n+1)]
=(n+1)^2*n^2
所以Sn=n^2*(n+1)^2/4
将n=100带入即可

Answer 4

1³+2³+3³+......+99³+100³

=(1+2+3+……+100)²
=5050²
=25502500

Answer 5

1+2+3+..+n = n(n+1)/2
n^2 = n(n+1) -n
=(1/3)[ n(n+1)(n+2)- (n-1)n(n+1) ] -n
1^2+2^2+3^2+..+n^2 = (1/3)n(n+1)(n+2) - n(n+1)/2
=(1/6)n(n+1)(2n+1)
n^3= n(n+1)(n+2) - 3n^2-2n
= (1/4)[ n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2) ] - 3n^2 -2n
1^3+2^3+3^3+..+n^3
= (1/4)n(n+1)(n+2)(n+3) - 3[(1/6)n(n+1)(2n+1)] -2[n(n+1)/2]
= (1/4)n(n+1)[ n^2+5n+6 - (4n+2) - 4]
=(1/4)n(n+1)(n^2+n)
= [n(n+1)/2]^2
put n=100
1³+2³+3³+......+99³+100³
=[100(101)/2]^2
= 25502500