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数学
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已知数列{an}的通元an=3n+1,求证:1、{an}是等差数列;2、若bn=pan+q(pq为常数)求证:
﹛bn﹜也是等差数列
人气:159 ℃ 时间:2020-02-05 10:58:39
解答
a[n+1]=3(n+1)+1=3n+4
a[n+1]-an =3n+4-(3n+1)=3
所以an是等差数列
bn=p(3n+1)+q
b[n+1]=p(3n+4)+q
b[n+1]-bn=p(3n+4)-p(3n+1)=3p
得证
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