求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标
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求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)
C(n,1)+2C(n,2)+...+nC(n,n) n是下标
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标
一,n为奇数,Sn=nC(n,n)+(1+n-1)C(n,1)+(2+n-2)C(n,2)+…+nC(n,n-1/2)=n[C(n,0)+C(n,1)+…+C(n,n-1/2)=n*2de(n-1)次方 二,n为偶数,Sn也一样
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标
等比数列前n项和SN=2(n-1)次方-c 求c
nSn+1-(n+1)Sn=n^2+cn两边同除以n(n+1)=>Sn+1/(n+1)-Sn/n=(n+c)/(n+1)S1,S2/2,S3/3成等差数列=>c=1为什么可以推得c=1 没看懂已知数列an的前项和为Sn,a1=1,nSn+1-(n+1)Sn=n^2+cn,S1,S2/2,S3/3成等差数列.(1)求C的值.(2)求数列{a
S n是a n的前n项和,且Sn+1=4a n+2,(n≥1) a1=1 1.b n=a n+1-a n,证明b n等比 2.c n=a n/2²,证明c n等差 3.求Sn
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C语言 n!求和问题输入n,计算Sn=1!+2!+3!+4!+.+n!的末6位. n
设数列的前n项和为sn,且Sn=n^2-6n+c(c∈R).
设数列an的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列设数列an的前n项和为Sn,已知S1=1,S(n+1)/Sn=(n+c)/n,且a1,a2,a3成等差数列求:1、求c的值2、求数列an的通项公式3、求人解答
an=n*2^n,求Sn
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证明:c(n,0)c(n,1)+c(n,1)c(n,2)+...c(n,n-1)c(n,n)=c(2n,n-1)
Sn=1/2n∧2+1/2n 求sn/s(n+1)
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数列{an}的前n项和为Sn,已知Sn=(n^2+3n)/2若数列{cn}满足c(n)=a(n)(n为奇数),c(n)=2^n(n为偶数),数列{cn}的前n项和为Tn,当n为偶数时,求Tn答案是Tn=((n^2+2n)/4)+((4/3)((2^n)-1)),
记数列{an}的前n项和Sn,且Sn=c/2*n^2+(1-c/2)n(c为常数,n属于N*),且a1,a2,a5成公比不等于1的等比数列(1)求c的值(2)设bn=1/anan+1,求数列{bn}的前n项和Tn
an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的
设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n若数列{bn}是首项为1,公比为c的等比数列,记A
已知数列{an}的前n项和为Sn,且Sn=(n²/2)+(11n/2).数列{bn}满足2b(n+1)=b(n+2)+bn.(n∈N*),且b3=11,b1+b2+.+b9=153.求数列{an}、{bn}的通项公式.说明:c右侧的n、n+1、n+2均为下标。