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已知数列{an},an属于自然数,sn=1/8(an+2)^2.求{an}是等差数列
人气:139 ℃ 时间:2020-05-05 06:26:11
解答
sn=1/8(an+2)^2
sn-1=sn=1/8(an-1+2)^2
8an=(an+2)^2-(an-1+2)^2=(an-an-1)(an+an-1+4)
=(an-an-1)(an+an-1)+4an-4an-1
4(an+an-1)=(an-an-1)(an+an-1)
an-an-1=4
{an}是等差数列
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