题目

1*3*5+3*5*7+.+9*11*13=多少,

提问时间：2020-08-07

答案

(2n-1)(2n+1)(2n+3)=(2n-1)(4n^2+8n+3)=8n^3+12n^2-2n-3
1+...+n=n(n+1)/2
1+...+n^2=(n+3n^2+2n^3)/6=n(n+1)(2n+1)/6
1+...+n^3=n^2(n+1)^2/4
当n=1,2,...,k时,
(2n-1)(2n+1)(2n+3)等于1*3*5,3*5*7,...,(2k-1)(2k+1)(2k+3)
k取5,求和就等于1*3*5+3*5*7+...+9*11*13
8(1+...+k^3)+12(1+...+k^2)-2(1+...+k)-3k
2k^2(k+1)^2+2k(k+1)(2k+1)-k(k+1)-3k
=(k+k^2)(2k+2k^2+4k+2-1)-3k
=(k+k^2)(2k^2+6k+1)-3k
=k(2k^3+8k^2+7k+1-3)
=k(2k^3+8k^2+7k-2)
=k(k+2)(2k^2+4k-1)
k取5,结果为
35*(70-1)=2415