Question

已知数列{an}的前n项和为Sn,对一切正整数,点（n,Sn）都在函数f(x)=2^(x+2)-4的图像上,1 求其通项

Answer 1

1、

根据题意,得

Sn=2^(n+2)-4=4(2^n-1)

a1=S1=4(2^1-1)=4

an=Sn-Sn-1=4(2^n-1)-4[2^(n-1)]=4[2^n-2^(n-1)]=2(2*2^n-2^n)=2^(n+1)

n=1时,同样成立.

{an}的通项公式为an=2^(n+1)

2.

bn=anlog2(an)

=2^(n+1)log2[2^(n+1)]

=(n+1)2^(n+1)

Tn=b1+b2+...+bn=2*2^2+3*2^3+...+n2^n+(n+1)2^(n+1)

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

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

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

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

=2+2^(n+1)-2-n2^(n+1)-2^(n+1)

=-n2^(n+1)

Tn/2=n2^(n+1)

Tn=n*2^(n+2)