Sn为前n项和,且满足2根号sn=an 1
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(1)∵2Sn=an2+n-4(n∈N*).∴2Sn+1=an+12+n+1-4.两式相减得2Sn+1-2Sn=an+12+n+1-4-(an2+n-4),即2an+1=an+12-an2+1,则an
Sn-S(n-2)=3(-1/2)^(n-1)an+an-1=3(-1/2)^(n-1)a1=s1=1a3+a2=3(-1/2)^2=3*1/4a5+a4=3(-1/2)^4=3*(1/4)^2:a2
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
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
(1)an+2Sn·S(n-1)=0(n≥2),又an=Sn-S(n-1)所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0
(1)∵数列a[n]的前n项和为S[n],且满足a[n]+2S[n]S[n-1]=0,n≥2∴S[n]-S[n-1]+2S[n]S[n-1]=0两边除以S[n]S[n-1],得:1/S[n-1]-1/
an+2Sn*S(n-1)=0而an=Sn-S(n-1)∴Sn-S(n-1)+2Sn*S(n-1)=0同除以Sn*S(n-1)整理:1/Sn-1/S(n-1)=2∴{1/Sn}为等差数列,公差2,首项
(1)∵Sn-Sn-1=2SnSn-1∴1Sn−1−1Sn=2即1Sn−1Sn−1=−2(常数)∴{1Sn}为等差数列  
1.2√Sn=an+14Sn=(an)^2+2an+14S1=(a1)^2+2a1+1=4a1,a1=14S(n-1)=[a(n-1)]^2+2a(n-1)+14an=4[sn-s(n-1)]=(an
1.等比数列an的前n项和An=(1/3)^n-c,a1=1/3-c,n>1时,an=An-A(n-1)=(1/3)^n-(1/3)^(n-1)=-2/3*(1/3)^(n-1)所以a1=-2/3,c
由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa
2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1
由等差数列的通项公式可得a2+a5+a17+a22b8+b10+b12+b16=2(2a1+21d)2(2b1+21d′)=a1+a22b1+b22=22(a1+a22)222(b1+b22)2=S2
1)an=1/2*(√Sn+√S(n-1))而:an=Sn-S(n-1)=[√Sn+√S(n-1)][√Sn-√[S(n-1)]所以:√Sn-√[S(n-1)]=1/2叠代加和,得:√Sn-√S1=(
应该是a1=0.5吧.(1)先把a1转化,Sn-(Sn-1)+2Sn*Sn-1=0,(Sn-1)-Sn=2Sn*Sn-1因为Sn不为0,所以两边同除Sn*Sn-1可得1/Sn-1/(Sn-1)=2很明
证明:(1)当n=1时左边=S1=a1=1右边=(2^1-1)/[2^(1-1)]=1左边=右边所以不等式成立(2)假设当n=k时等式成立即Sk=(2^k-1)/[2^(k-1)]那么当n=k+1时因
由题意可得S14T14=14(a1+a14)214(b1+b14)2=2a72b7=a7b7=3×14+24×14−5=4451,故答案为:4451.
2根号Sn=an+14Sn=an的平方+2an+14Sn_1=an_1的平方+2an_1+1〔n≥2〕又Sn-Sn_1=an所以4an=an的平方+2an-an_1的平方-2an_1划简为〔an+an
(1)证明:因为an=sn-s(n-1)所以有sn-s(n-1)+2sn*s(n-1)=0,即sn-s(n-1)=-2sn*s(n-1)同时除以2sn*s(n-1)整理得1/sn-1/s(n-1)=2
你的写法绝对有问题...害我走了很多弯路,以下[]表示下标b[n]-b[n-1]=(n+1)S[n]/n-nS[n-1]/(n-1)=(通分)=((n²-1)S[n]-n²S[n-