Question

an的前n项和Sn=n^2+2n,有两问,1,求an的通项.2,令bn=an*2^n,数列bn的前n项和.

Answer 1

根据第一问得到An=2n+1

Bn=An*2^n=(2n+1)*2^n

令Bn前n项和为Tn

则Tn=3*2^1+5*2^2+7*2^3+9*2^4+.+(2n-1)*2^n-1+(2n+1)*2^n (1)

2Tn=3*2^2+5*2^3+7*2^4+9*2^5+.+(2n-1)*2^n+(2n+1)*2^n+1 (2)

(1)-(2)

-Tn=3*2^1+2*2^2+2*2^3+2*2^4+2*2^5+.+2*2^n-(2n+1)*2^n+1

-Tn=2+2[2^1+2^2+2^3+2*4+.+2^n]-(2n+1)*2^n+1

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

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