早教吧作业答案频道 -->数学-->
数列bn满足bn=2的n次方+2n(n=1,2...)求数列bn的前n项和Tn
题目详情
数列bn满足bn=2的n次方+2n(n=1,2...)求数列bn的前n项和Tn
▼优质解答
答案和解析
数列bn的前n项和为:
Tn=2+2*1+2^2+2*2+……+2^n+2n
=(2+2^2+……+2^n)+(2+2*2+……+2n)
=2(1-2^n)/(1-2)+(2+2n)n/2
=2^(n+1)-2+n(n+1)
=2^(n+1)+n²+n-2
Tn=2+2*1+2^2+2*2+……+2^n+2n
=(2+2^2+……+2^n)+(2+2*2+……+2n)
=2(1-2^n)/(1-2)+(2+2n)n/2
=2^(n+1)-2+n(n+1)
=2^(n+1)+n²+n-2
看了数列bn满足bn=2的n次方+...的网友还看了以下:
已知数列{an}的前n项和为Sn,点(n,Sn/n)在直线y=1/2x+11/2上,数列{bn}满足 2020-03-30 …
已知数列{an}的前n项和为Sn,点(n,Sn/n)在直线y=1/2x+11/2上,数列{bn}满足 2020-03-30 …
已知数列an是各项均不为0的等差数列,Sn为其前n项和,且满足S2n-1=1/2an^2,数列bn 2020-04-09 …
设为数列的前n项和,对任意的,都有(m为常数,且m>0).(1)求证:数列是等比数列.(2)设数列 2020-05-13 …
已知数列{an}的前n项和为Sn,且满足Sn+n=2an(n∈N*).(1)证明:数列{an+1} 2020-05-13 …
数列{an}单调递增,满足a1=1,(an+1)四次方+(an)四次方+1=2[(an+1)²(a 2020-05-21 …
(2014•青岛一模)在数列{an}(n∈N*)中,其前n项和为Sn,满足2Sn=n-n2.(Ⅰ) 2020-06-12 …
数列{an}的前n项和Sn满足:Sn=n(n+3)/2(n属于N+)(1)设bn=2^n*an,求 2020-06-27 …
已知数列(An)满足A1=2,对于任意的n属于正整数,都有An大于0,且满足(n+1)×((An) 2020-07-20 …
已知数列{An}满足2a(n+1)=an+a(n+2),它的前n项和为An,且A3=5,A6=36. 2020-12-08 …