求2an+1-an=n+2的an通项,a1=0.5
问题描述：
求2an+1-an=n+2的an通项,a1=0.5
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)]...