求和1*4+2*5+3*6+...+n(n+3),已知1^2+2^2+3^2+...+n^2=(1/6)n(n+1)(2n+1).
问题描述:
求和1*4+2*5+3*6+...+n(n+3),已知1^2+2^2+3^2+...+n^2=(1/6)n(n+1)(2n+1).
答
1*4+2*5+3*6+...+n(n+3)=
1^2+2^2+3^2+...+n^2+
1*3+2*3+3*3+。。+3n
=(1/6)n(n+1)(2n+1)+[3-3^(n+1)]/(-2)
=(1/6)n(n+1)(2n+1)-[3-3^(n+1)]/2
答
难啊
答
1*4+2*5+3*6+...+n(n+3)=1^2+2^2+3^2+...+n^2+3(1+2+3+...+n)=n(n+1)(2n+1)/6+3n(n+1)/2=n(n+1)(n+5)/3