各項為立方和級數之級數之總和公式(15點)

Question

a1=13a2=13+23a3=13+23+33......an=13+23+33+.....n3則Sn=a1+a2+a3+......an=？(請以n表示)

Answer 1

an=[n(n+1)/2]^2=(n^4+2n^3+n^2)/4
Sn=(1/4)*Σ(n^4+2n^3+n^2)
=(1/4)*[(1/30)*n(n+1)(2n+1)(3n^2+3n-1)+(1/2)*n^2(n+1)^2+(1/6)*n(n+1)(2n+1)]
=(1/15)*n(n+1)(n+2)(3n^2+6n+1)

2006-05-07 14:04:05 補充：
抱歉，少乘上1/4正確答案(1/60)*n(n+1)(n+2)(3n^2+6n+1)

Answer 2

階差法(後項減前項)
0,S1,S2,S3,S4,S5
1,a2,a3,a4,a5
8,27,64,125
19,37,61
18,24
6

==> Sn=0C(n,0)+1C(n,1)+8C(n,2)+19C(n,3)+18C(n,4)+6C(n,5)

2006-05-09 20:41:06 補充：
前面多補兩個零, 再階差0,0,0,S1,S2,S30,0,1,a2,a30,1,8,271,7,196,126Sn=C(n+2,3)+6C(n+2,4)+6C(n+2,5)=n(n+1)(n+2)(4*5+6(n-1)*5+6(n-1)(n-2))/5!=n(n+1)(n+2)(20+30n-30+6nn-18n+12))/120=n(n+1)(n+2)(3nn+6n+1))/60

2006-05-09 21:27:54 補充：
According to Newton's Forward Difference Formula

2006-05-15 10:27:55 補充：
How do you know that
Σ (n^4)
=n(n+1)(2n+1)(3n^2+3n-1)/30 ?

2006-05-15 10:30:12 補充：
Do you know Σ (n^20)=?

Answer 3

a(1)=1^3，a(2)=1^3+2^3，a(3)=1^3+2^3+3^3
a(n)= 1^3+2^3+3^3+……+n^3
= sum (k^3) k從1到n
= sum (k*(k+1)*(k+2) – 3k*(k+1) + k) k從1到n
= sum (k*(k+1)*(k+2)) –3sum (k*(k+1)) + sum (k) k從1到n
= n(n+1)(n+2)(n+3)/4 - 3 n(n+1)(n+2)/3 + n(n+1)/2
= n^2(n+1)^2/4