an=2的n次方 2n-1

A1=1An=2An-1+2的n次方所以,A2=2*1+2^2=2+2^2A3=2*(2+2^2)+2^3=2^2+2*2^3A4=2*(2^2+2*2^3)+2^4=2^3+2*2*2^3+2^4=

an=n/(n-1)×a(n-1)+2n×3^(n-2)∴an/n=a(n-1)/(n-1)+2×3^(n-2)------(1)a(n-1)/(n-1)=a(n-2)/(n-2)+2×3^(n-3)

第23题没有看懂不过第1题我知道答案：an=3^(n-1)xan-1等式两边同时除以an-1变成an/an-1=3^(n-1)一次列出an-1/an-2=3^(n-2)an-2/an-3=3^(n-3

答案：lim[(1^n+2^n+3^n+4^n)]^(1/n)=lim[4^n*((1/4)^n+(2/4)^n+(3/4)^n+1)]^(1/n)=lim[4^n]^(1/n)*lim[(1/4)^

a(n+1)=2an+2^n,bn=an/2^(n-1),b(n+1)=a(n+1)/2^n,b1=a1/2^0=1a(n+1)/2^n=an/2^(n-1)+1,b(n+1)=bn+1,bn为首项为

(1)a(n+1)=2an+2^(n+1)等式两边同除以2^(n+1)a(n+1)/2^(n+1)=an/2ⁿ+1a(n+1)/2^(n+1)-an/2ⁿ=1,为定值a1/2=

采用Sn-q倍Sn,错位相减法!an=(2n-1)*(1/2)^nSn=1*(1/2)+3*(1/2)^2+5*(1/2)^3+……+(2n-1)*(1/2)^n0.5Sn=1*(1/2)^2+3*(

an+1-an=2^nan-an-1=2^n-1a2-a1=2^1-1an-a1=2^1+2^2+2^3+...2^n-1an=2^n+1

An=Sn-S(n-1)=2An+(-1)^n-2A(n-1)+(-1)^(n-1)=2An-2A(n-1)得An=2A(n-1)根据此式知道An为等比数列公比为2求第一项S1=2A1-1=A1得A1

若n=2kSn=(4+3(2k-1)+1)/2+2^k-2=2^k+3k-1=2^(n/2)+3n/2-1若n=2k+1Sn=2^k+3k-1+3(2k+1)+1=2^k+9k+3=2^((n-1)/

1.a_(1)=1,a_(n+1)=2a_(n)+2^(n)----------------1b_(n)=a_(n)/2^(n)将式子1左右两边同时除以2^(n+1),则:b_(n+1)=b_(n)+

当x=1时,an=2n-1,sn=n^2当x不等于1时,sn=a1+a2+a3+...+an=x+3x^2+5x^3+...(2n-1)x^nx*sn=x^2+3x^3+...(2n-3)x^n+(2

sn=3*n*(n+1)*(2n+1)/6+2*(1+n)*n/2+n=n*(n+1)*(2n+1)/2+n^2+2n=n^3+5/2n^2+5/2n

an=n(2^n-1)an=n*2^n-na1=1*2^1-1a2=2*2^2-2a3=3*3^3-3.an=n*2^n-nSn=a1+a2+a3+.+an=1*2^1-1+2*2^2-2+3*3^3

an=an-1+(n-1)+2^(n-1)∴an-an-1=n-1+2^(n-1)同理an-1-an-2=n-2+2^(n-2).a2-a1=1+2上述公式相加有an-a1=1+2+...+n-1+2

a(n+1)-an=2*3^n+1故有：a2-a1=2*3+1a3-a2=2*3^2+1...an-a(n-1)=2*3^(n-1)+1以上各式相加得：an-a1=2[3+3^2+..3^(n-1)]

（1）求a2,a3;a2=3^(2-1)+a1=3+1=4a3=3^(3-1)+a2=9+4=13(2）求证an=(3的n次方-1）/2an=3的n-1次方+an-1an-a(n-1)=3^(n-1)

证明：∵an+Sn=n²+2n-1,∴a(n+1)+S(n+1)=(n+1)²+2(n+1)-1则a(n+1)-an+S(n+1)-Sn=(n+1)²+2(n+1)-1-

我来试试吧.an=2^n+2n-1Sn=a1+a2+...+an=(2+2-1)+(4+4-1)+...+(2^n+2n-1)=(2+4+...+2^n)+(2+4+...+2n)-n=2[1-2^n

先给个答案.An=n(n-1)*[2^(n-2)]+[2^n-(-1)^n]/3.(n=1,2,3,4,...).原式两边同除以2^n,并设Bn=An/[2^(n-1)],则有B(n+1)=Bn+n+