设数列an为1,2x,3x2,4x3,...,nxn-1

S_n=n^2+n,S_(n-1)=〖(n-1)〗^2+n-1,∴a_n=S_n-S_(n-1)=2n (n>1),验证当n=1时,a_1=S_1=2,∴n=1时亦立,∴a_n=2n,

f(x)=x^n-x^(n+1)所以f'(x)=nx^(n-1)-(n+1)x^n所以f'(2)=n×2^(n-1)-(n+1)2^n=-(n+2)2^(n-1)所以an/(n+2)=-2^(n-1)

记Sn=a1+a2/2+a3/3+a4/4……+an/n=An+B,则a1=S1=A+B,当n>=2时,an/n=Sn-S(下标n-1)=An+B-[A(n-1)+B]=A,an=An,所以,an={

(1)根据题意,有An=(An-An-1)+(An-1-An-2)+…+（A2-A1）+A1=3-2^(2n-3)+3-2^(2n-5)+…+(3-2^3)+2再用分组求和法：=3n-【2^(2n-3

n=1,Sn=S1=a1X^2-a1*X-a1=0有一根为a1-1.有a1=1/2n=2,Sn=S2=a1+a2=1/2+a2X^2-a2*X-a2=0有一根为a1+a2-1=a2-1/2.有a2=1

∵点（an,sn）在直线y＝1/2（x2＋x）上∴Sn=1/2（an^2＋an）∴an=Sn-S(n-1)=1/2（an^2＋an）-1/2（a[n-1]^2＋a[n-1]）即1/2（an^2-an）

（1）当n=1时,a1>=3=1+2,an>=n+2成立；当n>1时,an=（an-1）^2-nan-1+1,令S=an-（n+2）=（an-1）^2-nan-1+1-（n+2）=（an-1）^2-(

a(n+1)-an=3*2^(2n-1)an-a(n-1)=3*2^(2n-3)...a3-a2=3*2^3a2-a1=3*2^1相加an-a1=3[2^1+2^3+2^5+2^7+...+2^(2n

(1)Sn/n=3n-2Sn=3n^2-2nn=1时a1=S1=1n≥1时an=Sn-S(n-1)=6n-5n=1时a1=1,成立∴an=6n-5(2)bn=3/[(6n-5)(6n+1)]=(1/2

n=1时,a1=S1=2a1+3-7,∴a1=4n>1时,Sn=2an+3n-7①,S(n-1)=2a(n-1)+3(n-1)-7②①-②得Sn-S(n-1)=2an+3n-7-[2a(n-1)+3(

1、a2=7a3=192、an+1=3an-2所以an+1-1=3(an-1)设bn=an-1则bn+1=3bn得证3、是求证吗?如果是求通项公式,那么由于a1=3,所以b1=2,则bn=2*3^(n

3an+1-3an=2即a(n+1)-an=2/3所以{an}是一个等差数列.故an=a1+(n-1)d=3+(n-1)*2/3所以,a100=3+(100-1)*2/3=69

3α-αβ+3β=?等什么?是1吗?

1.a(n+1)=3an-2则a(n+1)-1=3(an-1)令bn=an-1那么b1=2,b(n+1)/bn=3所以数列{bn}为等比数列,即数列{(an)-1}为等比数列2.b(n+1)=3bn=

方法一：A(n+1)-1=3An-3=3(An-1),且A1-1=2,所以数列{An-1}为公比为3,首项为2的等比数列方法二:设A(n+1)+k=3(an+k),即A(n+1)=3An+2k,则2k

Sn/n=3n-2Sn=3n^2-2nn>1时An=Sn-S(n-1)=6n-5n=1时A1=S1=1所以对一切n都有An=6n-5Bn=3/[(6n-5)(6n+1)]=(1/2)[1/(6n-5)

解1)Sn=1+2x+3x^2+4x^3+...+nx^(n-1)(1)xSn=x+2x^2+3x^3+...+(n-1)x^(n-1)+nx^n(2)(x≠1)(1)-(2)得(1-x)Sn=1+x

a(n+1)=an^2+6an+6=(an+3)^2-3,即a(n+1)+3=(an+3)^2,从而log5[a(n+1)+3]=2log5(an+3)而cn=log5(an+3),则结合上式即得c(

（1）由f(x)=（2x+3)/3x=2/3+1/x得出f(1/a(n-1))=2/3+a(n-1)=an(n>=2)所以an-a(n-1)=2/3即an为等差数列,公差为2/3由a2=f(1)=5/

1）当x≠1时令P=1+2x+3x^2+4x^3…+nx^(n-1)则xP=1x+2x^2+3x^3+4x^4…+nx^n故P-xP=1+x+x^2+x^3…+x^(n-1)-nx^n即（1-x）P=