sn=2 3 4 9 6 27
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1/S(n+1)=3/Sn+4令1/Sn=bn则有b(n+1)=3bn+4b(n+1)+2=3(bn+2)等比数列,则bn+2=(b1+2)*3^(n-1)b1=1/S1=1/a1=1所以bn=3^n
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
=10*a+a;这一步错了第二个a应该是初始值a你这样写全是变量b了#includeintmain(){\x09inta,n,b,c,Sn,i;\x09scanf("%d%d",&a,&n);c=a;
当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所
你说的应该是平方和的数列吧.解法如下:a(n)=n^2=n(n+1)-n,n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3,n=[n(n+1)-(n-1)n]/2,a(n)=[n(
∵1+12+14+…+(12)n-1=1−(12)n1−12=2−12n−1,∴Sn=2n−(1+12+122+…+12n−1)=2n-1−12n1−12=2n-2+12n−1.
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-
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
设首项为a1,公比为r,当r=1时,Sn=n(a1),此时Sn/S(n+1)的极限为1r≠1时,Sn=a1(1-r^n)/(1-r),Sn/S(n+1)=(1-r^n)/(1-r^(n+1)),极限为
①当n=1时,S1=1/2T1=1/2∴S1=T1②当n=2时,S2=1/2+1/12=7/12T1=1/3+1/4=7/12∴S2=T2③假定Sn=TnS(n+1)=Sn+1/(2n+1)-1/(2
An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(
由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所
再问: 再问:那个划横线的答案是不是错了再答:我觉得是
n>=2时:∵an=2Sn^2/[(2Sn)-1]∴Sn-(Sn-1)=2Sn^2/[(2Sn)-1]两边同时乘以(2Sn)-1并化简得2Sn(Sn-1)+Sn-(Sn-1)=0两边同时除以Sn(Sn
(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
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