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求2an+1-an=n+2的an通项,a1=0.5
人气:374 ℃ 时间:2020-08-20 04:53:16
解答
由2an+1 - an = n+2 可知2an - an-1 = n+1 ; a1= 0.5 a2 = 1.75两式相减可有:2(an+1 - an) - (an - an-1) = 1令bn+1 = an+1 - an,所以新数列bn 满足 bn+1 = (bn + 1)/2 ,b2= a2 - a1 = 1.25求出bn = b2/[2^(n-2)]...
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