设数列an为1,2x

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,

1、令x=0f(0+1)-f(0)=2×0+1f(1)=f(0)+1f(0)=0代入,得f(1)=1S1=f(1)=1a1=S1=12、将x换成x-1f(x-1+1)-f(x-1)=2(x-1)+1f

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

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)

Sn=2an-4Sn=2[Sn-S]-4Sn=2S+4Sn+4=2(S+4)所以Sn+4构成公比为2的等比数列Sn+4=(S1+4)*2^(n-1)利用S1=2a1-4=a1求出S1=a1=4Sn+4

t(1)=a(1)=1-a(1),a(1)=1/2=t(1).t(n)=1-a(n)a(n)=1-t(n)a(n+1)=1-t(n+1)a(n+1)t(n)=[1-t(n+1)]t(n)=t(n+1)

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

a1+|a2|+a3+|a4|=1+2+4+8=15

（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-(

（n+1)a^2n+1-nan^2+an+1an=0因式分解,得[a(n+1)+an]*[(n+1)a(n+1)-nan]=0数列{an}是首项为1的正数数列,所以a(n+1)+an>0,则(n+1)

(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

当n=1时,A(1)=S(1)=1^2=1=2*1-1；当n>1时,A(n)=S(n)-S(n-1)=n^2-(n-1)^2=2n-1.所以{A(n)}的通项公式是2n-1.因为B(n)=2*B(n-

a(n)=aq^(n-1),b(n)=na(1)+(n-1)a(2)+...+2a(n-1)+a(n),m=b(1)=a(1)=a,3m/2=b(2)=2a(1)+a(2)=2m+a(2),a(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(

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

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

/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)

(1)a(n+1)=an+2^n+1an=a(n-1)+2^(n-1)+1.a2=a1+2^1+1把上面n个等式相加得a(n+1)=2^n+2^(n-1)+.+2^1+n=2^(n+1)+n-2所以a

这次上图总算题目出对了.由于普通对数太难打了,我就先换底吧f(x)=lnx/ln2-ln4/lnx=lnx/ln2-2ln2/lnxf(2^an)=ln(2^an)/ln2-2ln2/ln(2^an)

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=