已知数列{an}a1=1 (1)若a(n+1)=an+n求通项公式an (2)若a(n+1)=an+2^(n-1)求通项公式an
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/13 14:17:00
xRN@|P~x^) /@LMhk*j^˘.kOR&\ dw0Mp>$=4[N0ɖHAub"$:]`"dP2t-sa[dic1(+ͱ-{x5"b$X_͓լO.7ӘMfL_":inql!oT"ppR62%i-/eX;՚|T)
|خj)
ec=b|X:cx_N
已知数列{an}a1=1 (1)若a(n+1)=an+n求通项公式an (2)若a(n+1)=an+2^(n-1)求通项公式an
已知数列{an}a1=1
(1)若a(n+1)=an+n求通项公式an
(2)若a(n+1)=an+2^(n-1)求通项公式an
已知数列{an}a1=1 (1)若a(n+1)=an+n求通项公式an (2)若a(n+1)=an+2^(n-1)求通项公式an
1)
a1 = 1
a2 = a1 + 1
a3 = a2 + 2
a4 = a3 + 3
……
an = a(n-1) + (n-1)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 1 + 2 + 3 + …… (n-1)
= 1 + [1+(n-1)]*(n-1)/2
= n*(n-1)/2 + 1
2)若a(n+1)=an+2^(n-1)求通项公式an
a1 = 1
a2 = a1 + 2^0
a3 = a2 + 2^1
a4 = a3 + 2^2
……
an= a(n-1) + 2^(n-2)
以上各式子相加,消去等式2端相同的项,最后残留
an = 1 + 2^0 + 2^1 + 2^2 + …… + 2^(n-2)
= 1 + 1*[2^(n-1)-1]/(2-1)
= 2^(n-1)
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
已知数列an满足a1=2,an=a(n-1)+2n,(n≥2),求an
已知数列{an}满足a(n+1)=an+n,a1=1,则an=
已知数列{an}中,a(n+1)=an+2^n,a1=3,求an
给定数列an={a1,a2,a3.an},bn=a(n+1)-an给定数列an={a1,a2,a3.an},bn=a(n+1)-an若数列bn为等差数列,则称数列an为二阶差数列,已知二阶差数列为an= {0,1,3,6...}求数列an与bn的通项公式
已知数列{an}满足a1=2,a(n+1)-an=a(n+1)*an,则a31=?
已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an
已知数列{an}满足a1=1,a(n+1)=nan n+1是角标
已知数列an,a1=2,且a(n+1)=2an+3n,求an
数列{an},已知a1=4,2a(n+1)=an+1(n∈N^*)求{an}通项公式.若a
已知数列an满足条件a1=-2 a(n+1)=2an/(1-an) 则an=
已知数列{An}满足a1=1,a(n+1)=2an+1 求证数列{an+1}是等比数列 求数列{an}通式
已知数列{an},A1=1 A(n+1)=2an/an+2 求a5
已知数列{an}满足a(n+1)=an+lg2,a1=1,求an
(高二数学)已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an(1)若a1,a3,a4成等比数列,求数列{an}的通项公式(2
【紧急--高一数学】已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an(1)若a1,a3,a4成等比数列,求数列{an}的通项
在数列{an}中,已知a1=2,若a(n+1)=an+2n(n为正整数) 求an
已知数列{an},a1=-1,a(n+1)=an+n,求通项公式