求14102035...的和值a(n)=n*(n+1)*(n+2)/6求S(n)即求下面公式的和值1+4+10+20+35+56+84+...+n*(n+1)*(n+2)/6

Question

求1 4 10 20 35...的和值
a(n)=n*(n+1)*(n+2)/6
求S(n)
即求下面公式的和值
1+4+10+20+35+56+84+...+n*(n+1)*(n+2)/6

Answer 1

化简a(n),a(n)=n^3/6 + n^2/2 + n/3
求S(n)实际上就是对n的立方、n的平方和n各项求和以后,带上系数再相加.
即S(n) = ∑a(n) = 1/6 *∑n^3 + 1/2 *∑n^2 + 1/3 *∑n (n^3表示n的立方.)
因为
∑n^3 = n^2*(n+1)^2/4
∑n^2 = n*(n+1)*(2n+1)/6
∑n = (1+n)*n/2
所以
S(n) = ∑a(n) = 1/6 *∑n^3 + 1/2 *∑n^2 + 1/3 *∑n
= 1/6 *[(n^2*(n+1)^2/4] + 1/2 *[n*(n+1)*(2n+1)/6] + 1/3 *[(1+n)*n/2]
= (n^4+6n^3+11n^2+6n)/24
= n*(n+1)*(n+2)*(n+3)/24