Sn是前n项和,记bn=nSn n的平方 c
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1=4b2=8d=4bn=4nnsn=3n-2(n+1)s(n+1)=3n+1a(n+1)=(3n+1)/(n+1)-(3n-2)/n所以an=(3n-2)/n-(3n-5)/(n-1)(n>=2)n
由nSn+1=2n(n+1)+(n+1)Sn,得(Sn+1)/(n+1)=2+Sn/n,{Sn/n}为等差数列,Sn/n=S1/1+(n-1)*2=2n+3,Sn=2n^2+3n,an=Sn-Sn-1
n三次方求和是((n*(n+1))/2)的平方,但是(-n)的三次方肯定不是,再说这道题目也没有要求(-n)的三次方啊,^是表示指数pf---平方a1=((a1+1)/2)pf,所以a1=1,Sn=(
(1)an+1=(n+2)/nSn,即S(n+1)-Sn=(n+2)/nSn,化简可得S(n+1)/(n+1)=2(Sn/n),即证得数列{Sn/n}是等比数列;(2)由(1)可知Sn=n*2^(n-
S(2n-1)=(A1+A(2n-1))×(2n-1)/2=(A1+A1+(2n-2)d)×(2n-1)/2=(A1+(n-1)d)×(2n-1)=An×(2n-1)同理T(2n-1)=Bn×(2n-
易知b1=4,b2=8,因此bn=4n,得4=sn+(n+2)/n*a(n)=sn+(n+2)/n*(sn-s(n-1)),因此sn=(n+2)/(2n+2)*s(n-1)+2n/(n+1),易用归纳
1.nS(n+1)-(n+1)Sn=n(n+c)两边同除n(n+1)S(n+1)/(n+1)-Sn/n=(n+c)/(n+1)S1/1,S2/2,S3/3是等差数列S(n+1)/(n+1)-Sn/n=
题目都打错了,这是江苏今年的高考题.
令Sn/n=bn则a(n+1)=Sn+2*bn,(n+1)*b(n+1)-n*bn=n*bn+2*bn,b(n+1)=2*bn故bn是等比数列第二问,相当于要求证明S(n+2)=4*S(n+1)-4S
an+sn=2n;a(n-1)+s(n-1)=2(n-1);上两式相减;得2an-a(n-1)=2;则2*(2-an)=(2-a(n-1));即2*bn=b(n-1);为等比数列;s1=a1,得a1=
我会我会Sn+1=Sn-2nSn+1Sn两边同除以Sn+1*Sn得1/Sn+1-1/Sn=2n以此类推1/Sn-1/Sn-1=2(n-1)1/Sn-1-1/Sn-2=2(n-2)...1/S2-1/S
1、A(n+1)=(n+2)sn/n=S(n+1)-Sn即nS(n+1)-nSn=(n+2)SnnS(n+1)=(n+2)Sn+nSnnS(n+1)=(2n+2)SnS(n+1)/(n+1)=2Sn/
n=n(n+1)=n^2+nSn=b1+b2+...+bn=(1^2+1)+(2^2+2)+...+(n^2+n)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)
证明:(1)注意到:a(n+1)=S(n+1)-S(n)代入已知第二条式子得:S(n+1)-S(n)=S(n)*(n+2)/nnS(n+1)-nS(n)=S(n)*(n+2)nS(n+1)=S(n)*
题目修改如下:已知数列{an}的前n项和为Sn,a1=1,nSn+1-(n+1)Sn=n²+n求annSn+1-(n+1)Sn=n²+n=n(n+1)两边同时除以n(n+1)Sn+
(1)证明:∵an+1=n+2nSn,∴Sn+1−Sn=n+2nSn∴Sn+1=2n+2nSn∴Sn+1n+1=2Snn∵a1=1,∴S11=1∴数列{Snn}是以1为首项,2为公比的等比数列;(2)
平方和的公式为S=n(n+1)(2n+1)/6所以,Sn=4×n(n+1)(2n+1)/6+4×n(n+1)/2=2n(n+1)(2n+1)/3+2n(n+1)=2n(n+1)(2n+4)/3=4n(
1=24bn=-3n+27≥03n≤27n≤9当n=9时bn}的前n项和Sn值最大b9=-3*9+27=0sn=(b1+b9)*9/2=(24+0)*9/2=54
等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1