数列an满足a1 1
若数列{an}满足a1=1,an+1=2an+n,则通项an= - - - .
解答:解:由an+1=2an+n,可得an+1+(n+2)=2(an+n+1),∴数列{an+n+1}是以a1+1+1=3为首项,2为公比的等比数列.∴an+n+1=3×2n 1,得到an=3×2n 1 n...
数列{an}满足a1=1,且对任意的m,n∈N*都有am+n=am+an+...
∵数列{an}满足a1=1,且对任意的m,n∈N*都有am+n=am+an+mn,∴an+1-an=1+n,∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n(...
已知数列an满足a1=1,an+1=3an - 2.写出an的前5项.猜想an...
a2=a3=a4=a5=1 所以,an=1 a(n+1)-1=3a(n)-3=3[a(n)-1]所以,【a(n+1)-1】÷【a(n)-1】=3 所以,a(n+1)-1 是等比数列,公比是3 ...
已知数列{an}满足a1=1,an>0,Sn是{an}的前n项和
an-a(n-1)=1/2 所以an为首相为1，公差为1/2的等差数列，即 an=1/2(n+1)记bn=an/2^n=(n+1)/2^(n+1)Tn=(1+1)/2^(1+1)+(1+2)/2^(1+2)+...
已知数列an满足a1=1,an+1=an/(an+1)(n∈N),求数列an的...
即1/a(n+1)-1/an=1 设bn=1/an，则b(n+1)=1/a（n+1），b1=1/a1=1 即b(n+1)-bn=1 则｛bn｝是等差数列，首项为b1=1，...
已知数列{an}满足:a1=1,an+1=an+1n为奇数2ann为偶数(n...
解答:解:(I)∵数列{an}满足:a1=1,an+1=an+1n为奇数2ann为偶数(n∈N*),设bn=a2n-1,∴b2=a3=2a2=2(a1+1)=4,b3=a5=2a4=2(a3+1)=10,同理,bn+1=a...
数列{an}满足a1=1,a2=2,an+2=2an+1 - an+2.(Ⅰ)设bn=an+...
(Ⅰ)由an+2=2an+1-an+2得,an+2-an+1=an+1-an+2,由bn=an+1-an得,bn+1=bn+2,即bn+1-bn=2,又b1=a2-a1=1,所以{bn}是首项为1,公差为2的等差数列....
设数列{an}满足:a1=1,an+1=3an属于N+。求{an}的通项公式及...
解：a(n+1)=3an a(n+1)/an=3为定值 所以{an}是以a1=1为首项，q=3为公比的等比数列 于是 an=a1xq^(n-1)=1x3^(n-1）=3^(n-1）Sn=a1(1-q^n)/(...
数列{an}满足a1=1,设该数列的前n项和为Sn,且Sn,Sn+1,2a1...
解析：由题意 2Sn+1=Sn+2a1=Sn+2 归纳法证明 当n=1时，S1=a1=1满足式子 假设n=k时，成立即Sk=(2k-1)/2k-1 则n=k+1时，Sk+1=1/2Sk+1=(2k-1)/2k...
数学题:数列an满足a1=1,n乘以an+1=(n+1),an+n乘以(n+1...
所以an/n是等差数列 a1/1=1 an/n=1+(n-1)*1=n an=n^2 bn=3^n*n b1 = 3*1 b2=3^2*2 Sn=b1+b2+...+bn =3*1+3^2*2+3^3*3+3^n*n (...