sn=2an-4n 1

Sn=(1/4)(an+1)^2S(n-1)=(1/4)[a(n-1)+1]^2相减且an=Sn-S(n-1),所以4an=(an+1)^2-[a(n-1)+1]^2[a(n-1)+1]^2=(an+

等比数列定义an+1=qanq不为零,且各项不为零等差数列定义an+1-an=pp为常数你上面提到的两个问题分别把{an-2an-1}、{an/2^n}看成an

1Sn=Sn-1+an-1+1/2an-an-1=1/2an=a1+(n-1)/2=1/4+(n-1)/2=n/2-1/423bn-bn-1=n3bn-3n/2-3/4=bn-1-(n-1)/2-1/

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

（1）证明：∵Sn-2an=2n，①∴Sn+1-2an+1=2（n+1）．②②-①，得：an+1-2an+1+2an=2，∴an+1=2an-2，∴an+1-2an-2=(2an-2)-2an-2=2

(2)a1=84(n+1)(Sn+1)=(n+2)^2.anSn+1=(n+2)^2.an/[4(n+1)](1)S(n-1)+1=(n+1)^2.a(n-1)/(4n)(2)(1)-(2)an=(n

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

n2=++n1先作n1=++n1,此时n1=n1+1=2+1=3,再作n2=n1=3n1=n2++先作n1=n2=3,再作n2=n2++=n2+1=3+1=4执行后n1=3,n2=4

1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=

Sn+1=4an+2Sn=4a（n-1）+2相减得Sn+1-Sn=4an+2-4a（n-1）-2an+1=4an-4a（n-1）an+1-2an=2(an-2an-1)bn=2bn-1(2)求数列{a

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/

你在步步高上看的题吧?前一阵子给人辅导做过这道题...这道题不是常规方法也用不了配凑系数出现新的等差等比数列这道题当时我们也研究了半天方法就是把a1,a2,a3,a4,...往后列,不要把a1=4带入

Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…

哎,看你着急的样子,我就替你解了此因果S(n+1)=4an+2Sn=4a(n-1)+2相减得：a(n+1)=4an-4a(n-1)移向得a（n+1）-2an=2(an-2a(n-1)){a(n+1)-

（1）当n=1时，a1＝s1＝14a21+12a1−34，解出a1=3，又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2（an-an

a(n+1)=4a(n)-3n+1,a(n+1)-(n+1)=4a(n)-4n=4[a(n)-n],{a(n)-n}是首项为a(1)-1=1,公比为4的等比数列a(n)-n=4^(n-1),a(n)=

(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

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.n=1时,S1=a1=(a1²+a1)/2,整理,得a1²-a1=0a1(a1-1)=0a1=0(与已知不符,舍去)或a1=1S1=a1=1n≥2时,Sn=(an²+

an+Sn=41a(n+1)+S(n+1)=2a(n+1)+Sn=422-1得2a(n+1)-an=0a(n+1)=1/2anan+Sn=4an≠0a(n+1)/an=1/2数列{an}是等比数列