Question

已知Sn是正项数列{an}的前n项和,且Sn=1/4an^2+1/2an-3/4(2)若an=2^nbn,求数列{bn}的前n项和

Answer 1

∵an=2^n *bn ∴bn=(2n+1)/2^n
∴{bn}的前n项和Tn=(2+1)/2+(2*2+1)/2^2+...+(2n+1)/2^n
=[2^n-1 *(2+1)+2^n-2 *(2*2+1)+...+2^0*(2n+1)]/2^n
=[(1*2^n+2*2^n-1 +3*2^n-2+...+(n-1)*2²+n*2)+(2^n-1 +2^n-2+...+2+2^0)]/2^n
令Un=1*2^n+2*2^n-1 +3*2^n-2+...+(n-1)*2²+n*2
则2Un=2^n+1 +2*2^n +3*2^n-1 +...+(n-1)*2³+n*2²
相减可得Un=2^n+1 +2^n +2^n-1 +...+2³+2² - 2n=4*(2^n-1)- 2n
∴{bn}前n项和Tn=[4*(2^n-1)- 2n+(2^n-1 +2^n-2+...+2+2^0)]/2^n=5 - (5+2n)/2^n

Answer 2

对不起手滑给上面的仁兄点赞点差了。。。手机党伤不起。。。也没法回复。。。所以只能这样道歉了