Question

已知数列{an}的首项a1=2/3，数列{1/an}是公比为1/2的等比数列，求数列[n/an}的前n项和

Answer 1

1/an=1/a1*q^(n-1)
1/an=1/(2/3)*(1/2)^(n-1)
1/an=3/2*(1/2)^(n-1)
1/an=3*(1/2)^n
an=1/3*(1/2)^n
an=2^n/3

1/a1=1/(2^1/3)=3/2^1
2/a2=2/(2^2/3)=6/2^2
3/a3=3/(2^3/3)=9/2^3
....
n/an=n/(2^n/3)=3n/2^n

S=3/2^1+6/2^2+9/2^3+....+3n/2^n

S/2=3/2^2+6/2^3+9/2^4+....+3(n-1)/2^n+3n/2^(n+1)

S-S/2
=3/2^1+3/2^2+3/2^3+....+3/2^n-3n/2^(n+1)
=3/2*[1-(1/2)^n]/(1-1/2)-3n/2^(n+1)
=3*[1-(1/2)^n]-3n/2^(n+1)
=3-3*(1/2)^n-3n*(1/2)^(n+1)
=3-3*(1/2)^n-3n/2*(1/2)^n
=3-(3+3n/2)*(1/2)^n
=3-(6+3n)/2*(1/2)^n
=3-(6+3n)*(1/2)^(n+1)

S=2*[3-(6+3n)*(1/2)^(n+1)]
S=6-3(2+n)*(1/2)^n