早教吧作业答案频道 -->数学-->
数列an的前n项和为Sn,Sn+an=-1/2n2-3/2n+1(n属于正自然数).设bn=an+n,证明数列bn是等比数列求数列{nbn}的前n项和Tn
题目详情
▼优质解答
答案和解析
Sn+an=-(1/2)n^2-(3/2)n+1
n=1
a1=-1/2
2Sn-S(n-1) = -(1/2)n^2-(3/2)n+1
2(Sn + (1/2)n^2 +(1/2)n -1) = S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1]=1/2
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S1 +(1/2)+(1/2) -1]=(1/2)^(n-1)
Sn + (1/2)n^2 +(1/2)n -1 = -(1/2)^n
Sn=1-n/2 -n^2/2 - (1/2)^n
an = Sn -S(n-1)
= -n +(1/2)^n
an +n = (1/2)^n
bn =an+n 是等比数列
nbn = n(1/2)^n
Tn =1b1+2b2+...+nbn
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=1/2
summation(i:1->n) i.(1/2)^(i-1)
= 4(n.(1/2)^(n+1) - (n+1).(1/2)^n + 1)
= 4[1- (n+2).(1/2)^(n+1)]
Tn =1b1+2b2+...+nbn
= (1/2)(summation(i:1->n) i.(1/2)^(i-1))
=2[1- (n+2).(1/2)^(n+1)]
n=1
a1=-1/2
2Sn-S(n-1) = -(1/2)n^2-(3/2)n+1
2(Sn + (1/2)n^2 +(1/2)n -1) = S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S(n-1) +(1/2)(n-1)^2+(1/2)(n-1) -1]=1/2
[(Sn + (1/2)n^2 +(1/2)n -1)]/[S1 +(1/2)+(1/2) -1]=(1/2)^(n-1)
Sn + (1/2)n^2 +(1/2)n -1 = -(1/2)^n
Sn=1-n/2 -n^2/2 - (1/2)^n
an = Sn -S(n-1)
= -n +(1/2)^n
an +n = (1/2)^n
bn =an+n 是等比数列
nbn = n(1/2)^n
Tn =1b1+2b2+...+nbn
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=1/2
summation(i:1->n) i.(1/2)^(i-1)
= 4(n.(1/2)^(n+1) - (n+1).(1/2)^n + 1)
= 4[1- (n+2).(1/2)^(n+1)]
Tn =1b1+2b2+...+nbn
= (1/2)(summation(i:1->n) i.(1/2)^(i-1))
=2[1- (n+2).(1/2)^(n+1)]
看了 数列an的前n项和为Sn,S...的网友还看了以下:
若n为一自然数,说明n(n+1)(n+2)(n+3)与1的和为一平方数n(n+1)(n+2)(n+ 2020-05-16 …
设a,m,n为自然数,a>1.证明若a^m+1|a^n+1,那么m|n设a,b,m,n为自然数,同 2020-05-16 …
若自然数n使得作竖式加法n+(n+1)+(n+2)均不产生进位现象,则称n为“可连数...若自然数 2020-05-16 …
在概率论中,为什么(n-1)S^2/ó^2是自由度为n-1的卡方分布?∑(xi-u)^2/σ^2∽ 2020-06-16 …
数理统计求助:为什么(n-1)S^2服从自由度为n-1的卡方分布为什么(n-1)S^2服从自由度为 2020-06-16 …
设A*,A^-1为阶方阵A的伴随阵、逆矩阵,则|A*A^-1|=设A*,A^-1为n阶方阵A的伴随 2020-06-18 …
对于一个自然数n,如果能找到自然数a和b(ab≠0),使n=a+b+ab,则称n是一个好数对于一个 2020-06-18 …
已知数列{an}的通项公式为log2[(n+1)/(n+2)],设其前n项和为Sn,则使Sn≮-5 2020-06-27 …
若9^n+C1(n+1)+...+C(n-1)(n+1)*9+Cn(n+1)是11的倍数,则自然数 2020-07-09 …
设X1,X2,……,Xn(n>2)为来自总体N(0,1)的简单随机样本,X为样本均值,记Yi=Xi 2020-07-21 …