> 数学 >
已知数列bn=2n(2的n次方-1),求数列(bn)的前n项的和
人气:239 ℃ 时间:2020-10-01 18:47:10
解答
let
S= 1.2^1+2.2^2+...+n.2^n (1)
2S= 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1)-(2+2^2+...+2^n)
=2n.2^(n+1)-2(2^n -1)
bn=2n(2^n -1)
= 2(n.2^n) - 2n
Tn = b1+b2+...+bn
=2S - n(n+1)
=4n.2^(n+1)-4(2^n -1) - n(n+1)
=(8n-4).2^n - (n^2+n-4)
推荐
猜你喜欢
© 2026 79432.Com All Rights Reserved.
电脑版|手机版