题目

关于高中数学的数列求和,怎么做?
已知an=2的n次方,bn=2n
求Tn=b1a1+b2a2+•••+bnan（b是分子,a是分母）

提问时间：2020-10-29

答案

没错,用错位相减法
Tn=2/2+(2*2)/(2^2)+(2*3)/(2^3)+...+2n/(2^n)------①
Tn/2=2/(2^2)+(2*2)/(2^3)+(2*3)/(2^4)+...+2n/(2^(n+1))-------②
①-②得
Tn/2=1+2/(2^2)+2/(2^3)+...+2/(2^n)-2n/(2^(n+1))
=1+2[1/(2^2)+1/(2^3)+...+1/(2^n)]-2n/(2^(n+1))
=1+2[2^(-2)+2^(-3)+...+2^(-n)]-2n/(2^(n+1))
=1+2[(1-(1/2)^n)-1/2]-2n/(2^(n+1))
=1+ 2(1-(1/2)^n)-1-2n/(2^(n+1))
= 2(1-(1/2)^n)-2n/(2^(n+1))
所以Tn=2[2(1-(1/2)^n)-2n/(2^(n+1))]
=4-(1/2)^(n-2)-4n/(2^(n+1))
=4-(1/2)^(n-2)-n/(2^(n-1))
哪里不理解的话可以追问