Question

数列求和的一个问题一个数列,有关系式 a1+2a2+3a3+4a4+5a5…nan=n(2n+1)(n+1),问前n项和

Answer 1

1) a1+2a2+3a3+4a4+5a5…nan=n(2n+1)(n+1)

2) a1+2a2+3a3+4a4+5a5...+(n-1)a(n-1)=(n-1)(2n-1)n

1)-2)得到

nan=6n^2

a(n)=6n

s(n)=6*(1+2+3+...+n)

=6*[n(n+1)/2]

=3n(n+1)

答案为B