an=4n^2-1分之一,求sn
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令Bn=1/An,则由已知可得:Bn-Bn-1=5;B1=1/3;则Bn为等差数列,所以Bn=1/3-(n-1)*5=16/3-5n=1/An所以An=1/(16/3-5n)
S(n+1)=4an+2;Sn=4a(n-1)+2a(n+1)=S(n+1)-Sn=4an-4a(n-1)a(n+1)=4[an-4a(n-1)]即:a(n+1)-2an=2[an-2a(n-1)][
S(n+1)=4an+2Sn=4a(n-1)+2,n≥2S(n+1)-Sn=a(n+1)=4an-4a(n-1)a(n+1)-2an=2(an-2a(n-1))令a(n+1)-2an=bnbn/b(n
因为Sn+1=4an+2Sn=4an-1+2故an+1=4an-4an-1an+1-2an=2(an-an-1)令an+1-2an=bn+1故bn+1=2bn所以bn=b2*2^(n-2)=(a2-2
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
a‹n›=1/[n(n+1)(n+2)]=(1/2)[1/n(n+1)-1/(n+1)(n+2)]=(1/2){[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}
a2=1-1/x=(x-1)/xa3=1-x/(x-1)=-1/(x-1)a4=1-(x-1)=x由此得周期为3a5=a2=1-1/xa2002=a1=xa2002+a2001+a2000=1+x-1
a(n+1)=3an+3n得a(n+1)+3n/2=3an+9n/2=3(an+3n/2){an+3n/2}以3为公比则{an+3n/2}前n项和为7(3^n-1)/4{3n/2}前n项和为3n(n+
∵a10=(a1+a19)/2∴S(2*9+1)=s19=(a1+a19)*19/2=a10*19∵b5=(b1+b9)/2∴T9=(b1+b9)*9/2=b5*9∵S19/T9=19/(9+4)=1
由题意得:2S(n+1)=4Sn+a1,则2Sn=4S(n-1)+a1解得:a(n+1)=2an,则{an}为等比数列,公比q=2所以,an=a1q^(n-1)=2^n同样:2S(n+1)=4Sn+a
2S(n+1)-Sn=22S(n+1)=Sn+22S(n+1)-4=Sn-2[S(n+1)-2]/(Sn-2)=1/2,为定值.S1-2=a1-2=1-2=-1,数列{Sn-2}是以-1为首项,1/2
an=Sn-Sn-1=-SnS(n-1)(Sn-Sn-1)/[SnS(n-1)]=-11/S(n-1)-1/Sn=-11/Sn-1/S(n-1)=1,为定值.1/S1=1/a1=1/(1/2)=2数列
对Sn=S(n-1)/[2S(n-1)+1]等式两边取倒数1/Sn=2+1/S(n-1)所以{1/Sn}是以2为公差,1/S1为首项的等差数列1/Sn=1/S1+2(n-1)=2n-1Sn=1/(2n
直接两边取倒数...得到:1/a(n+1)=(1/2)*(1/an)+1/2(如果我没理解错的话)接着把式子变形为1/a(n+1)-1=(1/2)*(1/an-1)这里就转化成{1/an-1}为一个等
显然可递推求出:因为sn+1/sn=an-2=sn-s(n-1)-2,所以有1/sn=-s(n-1)-2,进而有sn=1/[-s(n-1)-2],据s1=a1=-1/2,得出:s2=-2/3,进而反复
由S(n+1)/S(n)=(4n+2)/(n+1),可得a(n+1)/S(n)=S(n+1)/S(n)-1=(3n+1)/(n+1),所以S(n)=(n+1)/(3n+1)*a(n+1)以n-1代替n
此题采用裂项相消法1/(2+4)可化为1/2(1/2-1/4)所以题目=1/2[(1/2-1/4)+(1/4-1/6)+(1/6-1/8)+.+1/n-1/(n+1)]=1/2[1/2-1/(n+1)
a1=4>0,n≥2时,an的表达式为两算术平方根之和的一半,又算术平方根恒非负,因此{an}各项均非负,√Sn恒有意义.n≥2时,an=Sn-S(n-1)=[√Sn+√S(n-1)]/2[√Sn+√
1)累加法a1=2a2-a1=1/(1*2)a3-a2=1/(2*3)a4-a3=1/(3*4).an-a(n-1)=1/[(n-1)n]相加得an=2+(1-1/2)+(1/2-1/3)+(1/3-
a[n+1]=4a[n]-3n+1=4a[n]-4n+n+1因此a[n+1]-(n+1)=4a[n]-4n即b[n+1]=4b[n],也就是说b[n]是等比数列又b[1]=a[1]-1=1所以b[n]