已知递推关系求数列通项已知数列{an},a1=1,an+1=an+(2)(n≥1),求an.已知数列{an},a1=2,an+1=1/3an,求an.已知：a1=1,an+1=an+2n,求an.

Question

已知递推关系求数列通项

已知数列{an},a1=1,an+1=an+(2)(n≥1),求an.

已知数列{an},a1=2,an+1=1/3an,求an.

已知：a1=1,an+1=an+2n,求an.

Answer 1

(1)
a1=1,a=an+2
则,a-an=2
所以,an是以a1=1为首项,公差d=2的等差数列
则,an=a1+(n-1)d=1+(n-1)*2=2n-1
(2)
a1=2,a=(1/3)an
则,a/an=1/3
所以,an是以a1=2,公比q=1/3的等比数列
则,an=a1*q^(n-1)=2*(1/3)^(n-1)
(3)
a1=1,a=an+2n
所以,a-an=2n
则：
a1=1
a2-a1=2
a3-a2=4
……
an-a=2(n-1)
上述等式左右分别相加得到：an=1+2+4+……+2(n-1)
=1+2[1+2+3+……+(n-1)]
=1+2*[(1+n-1)*(n-1)/2]
=1+n(n-1)
=n²-n+1