已知正项等比数列an前n项和为sn,s2＝10,a3＝32设bn＝log2(an),T(2n)＝b1b2-b2b3+b3b4-b4b5+…+b(2n)b(2n+1)求T(2n)

Question

已知正项等比数列an前n项和为sn,s2＝10,a3＝32
设bn＝log2(an),T(2n)＝b1b2-b2b3+b3b4-b4b5+…+b(2n)b(2n+1)求T(2n)

Answer 1

由s2＝a1+a1q=10,a3＝a1q^2=32,
解得a1=2,q=4,所以an=2·4^(n-1)=2^(2n-1)
则bn=2n-1,所以 {bn}是等差数列,且d=2,
b1b2-b2b3=b2(b1-b3)= -2d·b2= -4·b2 ,b3b4-b4b5=b4(b3-b5)= -2d·b4= -4·b4,...
T(2n)＝b1b2-b2b3+b3b4-b4b5+…-b(2n)b(2n+1)
=-4(b2+b4+...+b2n)
=-4n(3+4n-1)/2
=-8n^2-4n