Question

很简单的数列题已知数列{an}an=2^(n-1),a1=1,设数列{n*an}的前n项和为Tn,求Tn.

Answer 1

Tn=1*2^0+2*2^1+3*2^2+,+n*2^(n-1)

2Tn= 2^1+2*2^2+,+(n-1)*2^(n-1)+n*2^n

-Tn=1+ 2^1+ 2^2+,+2^(n-1) -n*2^n

=1+2^n-2-n*2^n

=2^n*(1-n)-1

Tn=1-2^n*(1-n)