在Sn=na1 n(n-10) 2d,已知s2=5

(1)点（n,Sn/n）,（n∈N+）均在函数y=3x-2的图像上.那么Sn/n=3n-2∴Sn=3n^2-2n当n=1时,a1=S1=3-2=1当n≥2时,an=Sn-S(n-1)=(3n^2-2n

(1)an=sn-s(n-1)就有sn-s(n-1)+2sn*s(n-1)=0两边同除以sn*s(n-1)得1/sn-1/s(n-1)=2{1/sn}是等差数列1/sn=1/s1+(n-1)d=2n-

当然是证n=2上式成立问题在于,数学归纳法的道理你应该搞明白.数学归纳法一般用来证明与自然数有关的命题,它是基于归纳原理这一公理得出的（详见《初等数论》潘承洞潘承彪编第一章第一节）.数学归纳法使用起来

因为an,Sn,Sn-1/2成等比数列Sn（平方）=an*（Sn-1/2）由an=Sn-S（n-1）Sn（平方）=（Sn-S（n-1））*（Sn-1/2）化简得S（n-1）*Sn=S（n-1）/2-S

1.把（n,Sn/n)代入y=3x-2中化简得Sn=3n2-2nan=Sn-S(n-1)=6n-52.这一问以前做过,似乎要用放缩法吧..不记得了...

(Sn)²=[Sn-S(n-1)](Sn-1/2)(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2Sn+2SnS(n-1)-S(n-1)=0S(n-1

1)2Sn^2=an*(2Sn-1)=(Sn-S(n-1)(2Sn-1)2Sn^2=2Sn^2-2Sn*S(n-1)-Sn+S(n-1)S(n-1)-Sn=2Sn*S(n-1)两边同除以Sn*S(n-

an=Sn-Sn-1=4n+1(n>=2),a1=2*1+3=5,满足上式,an通项就是4n+1,即证实等差数列

a1=S1=3+2=5Sn=3n²+2n①S(n-1)=3(n-1)²+2(n-1)②①-②得an=6n-1d=an-a(n-1)=6n-1-6(n-1)+1=6

Sn=3n的平方+2nSn-1=3(n-1)^2+2(n-1)An=Sn-Sn-1=3n^2+2n-3(n-1)^2-2(n-1)=3n^2+2n-3n^2+6n-3-2n+2=6n-1

结论：(1)c=3(2)a[n]=3*2^n-3.注：[]内为下标1.S[n]=2a[n]-3n,当n=1时可得a[1]=32.当n>=2时S[n]=2a[n]-3n且S[n-1]=2a[n-1]-3

an=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2,a1=1/6所以S1=a1=1/6n>=2时,Sn=a1+a2+...+an=[1/1*2-1/2*3]/2+[1

竟然第一问会做,那么an=n就不给你说了bn=np^(n)那么Tn=p+2p^2+3p^3+.np^n①则pTn=p^2+2p^3+.(n-1)p^n+np^(n+1)②由①-②得(1-p)Tn=p+

/>an=sn-s(n-1)=an+n^2-n-[a(n-1)+(n-1)^2-(n-1)]=an-a(n-1)+2n-2a(n-1)=2(n-1)an=2n所以,数列an为公差为2的等差数列.

1)由题意知,a(n+1)=2Sn+1Sn=(a(n+1)-1)/2S(n-1)=(a(n)-1)/2两式左右分别相减,化简后得到a(n+1)=3a(n)a1=t,a2=2t+1a2=3a1=>t=1

将a[n+1]=S[n+1]-S[n]代人得到：S[n]=4(S[n+1]-S[n])+14S[n+1]=5S[n]-14(S[n+1]-1)=5(S[n]-1)(S[n+1]-1)/(S[n]-1)

 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

Sn=3n^2-2n当n>=2,sn-1=3(n-1)^2-2(n-1)an=sn-sn-1=3n^2-2n-3(n-1)^2+2(n-1)=6n-5(n>=2)当n=1,s1=a1=1满足上个式子所