Question

高考数学题目

已知数列《an》通向公式an=4^n - 2^n，前n项和为Sn,则数列《2^n / Sn》前n项和Tn等于多少？

Answer 1

解:
Sn=4(4^n-1)/(4-1)-2(2^n-1)/(2-1)
=[4^(n+1)-4)/3-[2^(n+1)-2]
=[4^(n+1)-4-3*2^(n+1)+6]/3
=[2^(n+1)*2^(n+1)-3*2^(n+1)+2]/3
=[2^(n+1)-1][2^(n+1)-2]/3
2^n/Sn
=3*2^n/[2^(n+1)-1][2^(n+1)-2]
=3/2*2^(n+1){1/[2^(n+1)-2]-1/[2^(n+1)-1]}
=3/2*2^(n+1)/[2^(n+1)-2]-3/2*2^(n+1)/[2^(n+1)-1]
=3/2*{1+2/[2^(n+1)-2]}-3/2*{1+1/[2^(n+1)-1]}
=3/2{2/[2^(n+1)-2]-1/[2^(n+1)-1]}
=3/2{1/(2^n-1)-1/[2^(n+1)-1]}
所以
Tn
=3/2{1-1/3+1/3-1/7+1/7-1/15+...+1/(2^n-1)-1/[2^(n+1)-1]}
=3/2{1-1/[2^(n+1)-1]}
=3/2-3/[2^(n+2)-2]