a1+3an
数列{an}中,an+1=an1+3an,a1=2,则a4=
∵an+1=an1+3an，a1=2，∴a2=21+3×2=27，a3＝271+3×27=27+6＝213，∴a4=2131+3×213=213+6＝219．故答案为：219．
请问在数列{a,}中, a1=2, an+1=3an+1, 则a4=多少?
a1+1 = 3a2 +1 => 3a2 = 2 => a2 = 2/3 a2+2 = 3a3+2 => 3a3 =2/3 => a3 = 2/9 a3 = 3a4 => 3a4 = 2/9 => a4 = 2/27 则a4 =...
已知数列{an}满足an+1=3an+3n,且a1=1,求an.
由an+1=3an+3n,得an+1+x(n+1)+y=3(an+xn+y),即an+1=3an+3xn+3y-x(n+1)-y=3an+2xn+2y-x.类比an+1=3an+3n,只需2xn+2y-x=3n2x=3,2y-x=0,x...
a1=2,an+1=an/1+3an求an
因为an+1=an/1+3an所以1/a(n+1)=1/an+31/a(n+1)-1/an=3所以1/an是以公差为3，首项为1/a1=1/2的等差数列所以1/an=1/2+3（n-1）=3n-5/2所以an...
数列{an}中,a1=1,an+1=3an+2,则通项an= - - - .
解答:解:设an+1+k=3(an+k),得an+1=3an+2k,与an+1=3an+2比较得k=1,∴原递推式可变为an+1+1=3(an+1),∴an+1+1an+1=3,∴{an+1}是一个以a1+1=2...
数列{an}中,an+1=an1+3an,a1=2,则a4等于( )A.165B.219C...
由题意可得，1an+1＝ 1+3anan＝1an+3即1an+1?1an＝3∵1a1＝12∴数列{1an}是以12为首项，以3为公差的等差数列∴1an＝12+3(n?1)∴a4＝219故选：B ...
已知数列{an}满足a1=1,an+1=3an+2,则数列{an}的通项...
由an+1=3an+2,得an+1+1=3(an+1),又a1=1,所以{an+1}是以2为首项、3为公比的等比数列,∴an+1=2×3n 1,an=2×3n 1-1.故答案为:2×3n-1-1.
已知数列{an}满足a1=1,an+1=3an,求{an}的通项公式及前n项...
an+1=3an得an+1/an=3也就是说an为a1=1，公比为3的等比数列 所以an=a1×q^(n-1)=3^(n-1)Sn=a1(1-q^n)/(1-q) =(1-3^n）/(1-3) =（3^n-1...
在数列{an}中,an+1=3an+2n+1,a1=1,求数列{an}的通项...
∵an+1=3an+2n+1,∴an+1+(n+2)=3[an+(n+1)],又∵a1+2=1+2=3,∴数列{an+(n+1)}是以首项、公比均为3的等比数列,∴an+(n+1)=3n,∴an=3n-n-1....
a1=5,an+1=3an求an的通项公式
a1=5,a(n+1)=3an 说明数列{an}是等比数列,公比是q=3 所以an=a1*q^(n-1)=5*3^(n-1)如果不懂,请Hi我,祝学习愉快!