Question

已知等差数列（an）的前n项和为Sn,且a3=5,S15=225,设bn=2^a n+2n,求数

Answer 1

(1)

S15=225

15(a1+a15)/2=15a8=225

a8=15

a3=5

所d=(15-5)/5=2

故a1=a3-2d=5-2*2=1

所an=a1+(n-1)d=1+2(n-1)=2n-1

(2)

bn=2^an+2n=2^(2n-1)+2n

分组求和

Tn=b1+b2+...+bn

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

=(2+2^3+...+2^(2n-1))+2(1+2+...+n)

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

=2*(4^n-1)/3+n(n+1)

望采纳~~~