sn=-3 2^2 205 2n通项公式
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[基本原理:其实所有数列求和的方法基本都一样:裂项求和.这个数列也不例外.所谓裂项求和,就是先裂项,再求和.求和过程中必会相互消去,从而简化.]解:(裂项)由已知易得:an=1/(2n-1)(2n+1
Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3
an=2^(n)-1-(2^(n-1)-1)=2*(2^(n-1))-1-2^(n-1)+1=2^(n-1)你上面少个-1
当n=1时,a1=S1=32-1=31.当n≥2时,an=Sn-Sn-1=32n-n2-[32(n-1)-(n-1)2]=33-2n.当n=1时,上式也成立.∴an=33-2n.令an≥0,解得n≤3
平方和公式n(n+1)(2n+1)/6即1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6(注:N^2=N的平方)证明1+4+9+…+n^2=N(N+1)(2N+1)/6证法一(归纳猜想
Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b共m项A(n-m+1)=A1+(n-m)dA(n-m+2)=A2+(n-m)d……A(n-m+m)=An=Am+(n-
S2n/Sn=(4n+2)/(n+1),即S2/S1=6/2S2=3S1=3a1=3a2=S2-a1=3-2=1d=2-1=1an=n
sn=32n-n^2=33n-2*(n+1)n/2=[33-2]+[33-4]+...+[33-2n]∴an=33-2nTn=|a1|+|a2|+...|an|=(|33-2|+|33-4|+...+
1.n=1时,a1=S1=1²+1=2n≥2时,Sn=n²+nS(n-1)=(n-1)²+(n-1)an=Sn-S(n-1)=n²+n-(n-1)²-
2a[n]-n-1=a[n-1]【1】待定系数:2(a[n]+xn+y)=a[n-1]+x(n-1)+y【2】将【1】式a[n-1]代入上式:(注意:也可变换后用a[n]代入上式,看方便确定)2(a[
1.Sn×S(n-1)=2An=2(Sn-S(n-1))两边除以Sn×S(n-1)1=2(Sn-S(n-1))/(Sn×S(n-1))1/S(n-1)-1/Sn=1/21/Sn-1/S(n-1)=-1
(1)令n=1a1=S1=32-1+1=32Sn=32n-n²+1Sn-1=32(n-1)-(n-1)²+1an=Sn-Sn-1=32n-n²+1-32(n-1)+(n-
由1/S1+1/S2+1/S3+.+1/Sn=n/(n+1),知,当n=1时,s1=2,当n≥2时1/S1+1/S2+1/S3+.+1/Sn-1=(n-1)/n,两式相减得,1/sn=1/[n(n+1
Sn^2-n^2×Sn-(n^2+1)=0(Sn+1)[Sn-(n^2+1)]=0数列各项为非零实数,S1≠0,且Sn不恒为0,因此只有Sn=n^2+1n=1时,a1=S1=1+1=2n≥2时,an=
再问: 再问:那个划横线的答案是不是错了再答:我觉得是
f(n)=[1/2(n+1)n]/[(n+32)(n+2)(n+1)1/2]=n/(n+32)(n+2)=n/(n^2+34n+64),f(n)×(n/n)=1/[n+(64/n)+34]且n为正整数
已知数列a‹n›首相a₁=3,通项a‹n›和前n项和S‹n›之间满足2a‹n›=S̸
An=Sn-Sn-1由2An=Sn*Sn-1(n>2)得2Sn-2Sn-1=Sn*Sn-1简单说明Sn,Sn-1不为0(反证法)同除Sn*Sn-1:2/Sn-1-2/Sn=1令Bn=2/Sn则Bn-B
(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1
错位相减Sn=1*3+3*3^2+5*3^3+.+(2n-3)*3^(n-1)+(2n-1)*3^n(1)同乘以33Sn=1*3^2+3*3^3+.+(2n-3)*3^n+(2n-1)*3^(n+1)