Question

求数列an=n·2^2n 的前n项和Tn

Answer 1

Tn=1*2^2+2*2^4+...+n*2^2n

4Tn=1*2^4+2*2^6+...+n*2^(2n+2)

下式减上式

3Tn=n*2^(2n+2)-(2^2+2^4+2^6+...+2^2n)

=n*2^(2n+2)-4*(1-2^2n)/(1-4)

=[(3n-1)*4^(n+1)+4]/3

故Tn=[(3n-1)*4^(n+1)+4]/9