Question

在等差数列{an}中,若a1+2a2+3a3+L+nan=n(n+1)(n+2),则an=

Answer 1

an为等差数列，所以
an=a1+(n-1)d

则
a1+2a2+3a3+...+nan=a1+2(a1+d)+3(a1+3d)+...+n(a1+(n-1)d)
=a1(1+2+...+n）+2*d+3*2d+....+n(n-1)d
=n(n+1)a1/2+d*(∑n(n-1))
=n(n+1)a1/2+d*(∑n^2-∑n)
=n(n+1)a1/2+n(n+1)(2n+1)d/6-n(n+1）d/2
=n(n+1)[a1/2+(2n+1)d/6-d/2)

a1+2a2+3a3+L+nan=n(n+1)(n+2)

所以
a1/2+(2n+1)d/6-d/2=n+2
所以
a1/2+d/6-d/2=2
2nd/6=n

d=3
a1=6

an=6+(n-1)*3=3n+3

Answer 2

3N+3