An=n[100-(2n-1)]
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1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x
an/a(n-1)=(n+1)/(n-1)(n>=2)a(n-1)/a(n-2)=n(n-2)...a2/a1=3/1上式全部相乘an/a1=(n+1)!/2(n-1)!=n(n+1)/2,an=n(
(1)bn=a(2n+1)+4n-2b(n+1)=a(2n+3)+4(n+1)-2=a(2n+2+1)+4n+2=a(2n+2)-2(2n+2)+4n+2=a(2n+1+1)-2(2n+2)+4n+2
A(n+1)/An=(n+2)/nAn/A(n-1)=(n+1)/(n-1)A(n-1)/A(n-2)=n/(n-2).A3/A2=4/2A2/A1=3/1把所有式子的左边相乘,右边相乘,等式仍成立.
(1)由已知a2=2a1+2,a3=2a2+3=4a1+7,若{an}是等差数列,则2a2=a1+a3,即4a1+4=5a1+7,得a1=-3,a2=-4,故d=-1. &nbs
不知道你的题目是不是这样
An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)则An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+1/(2n+1)+1/(2n+2)则An+1-A
(N+1)是下标么?5对什么,看不太懂
这是斐波那契数列:1、1、2、3、5、8、13、21、34、55、89、144该数列从第12项,满足an>100n的最小值是12请点击下面的【选为满意回答】按钮,
C(k,n)ak=n!/((n-k)!*k!)*(k(k+1))/2=(n-1)!/((n-k)!(k-1)!)*(n(k+1))/2=C(k-1,n-1)*n/2*(k+1)An=n/2*[C(0,
1.an=-a(n-1)-2n+1an+n=-a(n-1)-n+1=-[a(n-1)+(n-1)](an+n)/[a(n-1)+(n-1)]=-1,为定值.a1+1=3+1=4数列{an+n}是以4为
假设An=1/n(n+1)成立当n=1时A1=1/2成立令n=k(k>=0)时Ak=1/k(k+1)成立当n=k+1A1+A2+…+Ak+A(k+1)=k^2*Ak+A(k+1)=(k+1)^2*A(
a(n+1)=2an-n^2+3n=2an+(n+1)^2-(n+1)-2n^2+2n将(n+1)^2-(n+1)移过去得a(n+1)-(n+1)^2+(n+1)=2(an-n^2+n)再两边同除(a
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
应该是A(n+1)=An+2n吧~~~=>a(n+1)-an=2n所以an-a(n-1)=2(n-1)a(n-1)-a(n-2)=2(n-2)...a2-a1=2*1把左边加起来,右边加起来得到an-
【方法1:强行展开a(n)表达式】1+2+……+n=n(n+1)/21^2+2^2+……+n^2=n(n+1)(2n+1)/61^3+2^3+……+n^3=n^2(n+1)^2/41^4+2^4+……
待定系数法因为a(n+1)=2an-n^2+3n设a(n+1)+p(n+1)^2+q(n+1)=2(an+pn^2+qn)展开整理得a(n+1)=2an+pn^2+(q-2p)-(p+q)与原式一一对
an=(n+1)(n+2)再问:有木有过程?再答:原式整理后得到an=(n+1)(an-1/n+1)试值:a2=(2+1)(6/2+1)=(2+1)(2x3/2+1)=12=3x4a3=(3+1)(1
2a(n+1)-an=n-2/n(n+1)(n+2)2a(n+1)-2/(n+1)(n+2)=an-1/n(n+1)[a(n+1)-1/(n+1)(n+2)]/[an-1/n(n+1)]=1/2bn=
(1)a(n+1)/2^(n+1)=an/(an+2^n)2^(n+1)/a(n+1)=(an+2^n)/an=1+2^n/an2^(n+1)/a(n+1)-2^n/an=1所以{2^n/an}是以公