Question

已知等比数列an的前n项和为Sn=-a*2^n+b,且a1=3

（1）求a，b的值及数列an的通项公式
（2）设bn=n/an，求数列bn的前n项和Tn

Answer 1

a1=S1=-2a+b=3
an=Sn-S(n-1)=-a*2^n+b-[-a*2^(n-1)+b]=-a*2^(n-1)
a1=-a*2^0=3
a=-3
b=3+2a=-3
an=3*2^(n-1)
bn=n/an=n/[3*2^(n-1)]
Tn=1/(3*1)+2/(3*2)+3/(3*2^2)+...+n/[3*2^(n-1)]
1/2Tn=1/(3*2)+2/(3*2^2)+3/(3*2^4)+...+n/(3*2^n)
Tn-1/2Tn=1/3+1/(3*2)+1/(3*2^2)+1/[3*2^(n-1)]-n/(3*2^n)
1/2Tn=1/3*(1-1/2^n)(1-1/2)-n/(3*2^n)
Tn=4/3-(n+2)/[3*2^(n+1)]

Answer 2

a=-3,b=-3,an=3*2^n