Question

对于数列an=n^3 求前N项和Sn=......

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Answer 1

1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2
证明：
(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]
=(2n^2+2n+1)(2n+1)
=4n^3+6n^2+4n+1

2^4-1^4=4*1^3+6*1^2+4*1+1
3^4-2^4=4*2^3+6*2^2+4*2+1
4^4-3^4=4*3^3+6*3^2+4*3+1
......
(n+1)^4-n^4=4*n^3+6*n^2+4*n+1

各式相加有
(n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n

4*(1^3+2^3+3^3+...+n^3)=(n+1)^4-1+6*[n(n+1)(2n+1)/6]+4*[(1+n)n/2]+n=[n(n+1)]^2

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

Answer 2

∑n³=（∑n）²，具体可由数学递推法证明
所以Sn=（∑n）²=（n（n+1）/2）²

Answer 3

Sn = n² (n+1)²/4

Answer 4

(n+1)^4-n^4=4n^3+6n^2+4n+1.
∴n^3=(1/4)[(n+1)^4-n^4]-(3/2)n^2-n-1/4
∴左边=∑i^3=(1/4)[(n+1)^4-1]-(3/2)*(1/6)n(n+1)(2n+1)-(1/4)n-(n+1)n/2
=(1/4)(n^4+4n^3+6n^2+4n-2n^3-3n^2-n-n)-(1/2)(n^2+n)
=(1/4)(n^4+2n^3+n^2)
=[(1/2)n(n+1)]^2

Answer 5

Sn=n²(n+1)²/4