1.求Sn=2/2+3/2∧2+4/2∧3+...+n/2∧n-1+n+1/2∧n2.求数列1,3+5,7+9+11,13+15+17+19,.的前N项和.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 10:32:21
xRMJ1& /R
3s7,A/ bu#R7
Auת+ˋS-..$/%It{M`~\iy6 ˵0S}=8@+jC#4-6`'zxv12VX1o?Uf?_.'5TWHxV<305DYYϏ1ñQ<FA=||ǘ%goG[b
,'|gwR]SQߟ'U Q5j>-t䆻OYx
D픛6
1.求Sn=2/2+3/2∧2+4/2∧3+...+n/2∧n-1+n+1/2∧n2.求数列1,3+5,7+9+11,13+15+17+19,.的前N项和.
1.求Sn=2/2+3/2∧2+4/2∧3+...+n/2∧n-1+n+1/2∧n
2.求数列1,3+5,7+9+11,13+15+17+19,.的前N项和.
1.求Sn=2/2+3/2∧2+4/2∧3+...+n/2∧n-1+n+1/2∧n2.求数列1,3+5,7+9+11,13+15+17+19,.的前N项和.
1.Sn/2=2/2^2+3/2^3+...+(n+1)/2^(n+1)
=3/2∧2+4/2∧3+...+n/2∧n-1+n+2/2∧(n+1)-[1/2^2+1/2^3+1/2^4+...+1/2^(n+1)]
=S(n+1)-2/2-[1/2-1/2^(n+1)]
=Sn+(n+2)/2^(n+1)-3/2+1/2^(n+1)
=Sn+(n+3)/2^(n+1)-3/2
因此Sn/2=3/2-(n+3)/2^(n+1)
Sn=3-(n+3)/2^n
2.前n项包括的奇数的个数为:1+2+3+...+n=n(n+1)/2
因为前n个奇数之和是n^2,所以数列1,3+5,7+9+11,13+15+17+19,.的前N项和为:
[n(n+1)/2]^2=n^2(n+1)^2/4
a1=1.sn=2an+1.求sn
2Sn+Sn-1=3-8/2^n,求Sn
高二数学数列求和问题!在数列an中,a1=1,Sn为an前n项和,an=S(n-1) 求Sn,an请帮我看看我分别先算Sn和an哪里出错了?1.先算Sn因为an=2Sn-1 且an=Sn-Sn-1所以 Sn-Sn-1=2Sn-1Sn=3Sn-1所以Sn是一个等比数列 公比为3 运
Sn=1/2n∧2+1/2n 求sn/s(n+1)
1+2+3+4+.+n,求Sn
sn=2*3^n,求an
a1=1,Sn=2an+1求Sn
数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn
已知数列前n项和为Sn,且Sn=-2n+3,求an及Sn
数列bn=2^n(4n-3),求Sn
Sn=1+2+4+...+2的N次方,求Sn
sn=1+2+4+...+2^(n-1),求sn
数列求和:sn=1+1/2+1/3+…+1/n,求sn
Sn为数列{an}前n项和,(2n-1)Sn+1-(2n+1)Sn=-4n-3 ,求{an}通项公式
an=n*2^n,求Sn
an=2^n+n,求Sn
已知Sn是数列an的前n项和,an的通向公式为2n 设Tn=(Sn/Sn+1) +( Sn+1/Sn)-2设数列{an}的前项和为sn,a1=2,点(Sn+1,Sn)在直线(X/n+1)-(y/n)=1(n是正整数,1.求an的通项公式;2 .设Tn=(Sn/Sn+1) +( Sn+1/Sn
Sn=2/2+3/2²+4/2³+····+(n+1)/(2^n) 求Sn.