设数列{an}满足:a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.谢拉...急ING~
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![设数列{an}满足:a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.谢拉...急ING~](/uploads/image/z/1765604-20-4.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D1%2Ca2%3D5%2F3%2Can%2B2%3D5%2F3an%2B1-2%2F3an%28n%3D1%2C2%2C3%2C...%29%281%29%E4%BB%A4bn%3Dan%2B1-an%28n%3D1%2C2%2C3%2C...%29%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%EF%BC%8E%E8%B0%A2%E6%8B%89%EF%BC%8E%EF%BC%8E%EF%BC%8E%E6%80%A5ING%7E)
设数列{an}满足:a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.谢拉...急ING~
设数列{an}满足:a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)
(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;
(2)求数列{an}的前n项和Sn.
谢拉...急ING~
设数列{an}满足:a1=1,a2=5/3,an+2=5/3an+1-2/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.谢拉...急ING~
由an+2=5/3an+1-2/3an
可得an+2-an+1=2/3an+1-2/3an
bn+1=2bn/3
bn+1/bn=2/3
{bn}是公比为2/3的等比数列
b1=a2-a1=2/3
bn=(2/3)^n
设Sn为{bn}前n项和
Sn=2[1-(2/3)^n]=a2-a1+a3-a2+a4-a3……+an+1-an
=an+1-a1=an+1-1
an+1=2[1-(2/3)^n]+1
an=2[1-(2/3)^(n-1)]+1
an=3-2(2/3)^(n-1)
数列{an}的前n项和Sn
Sn=3n-2[(2/3)^0+(2/3)^1+……+(2/3)^(n-1)]
Sn=3n-6+4(2/3)^(n-2)
计算结果也许有小问题,思路绝对正确