求2an+1-an=n+2的an通项,a1=0.5

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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)] + (n-2)/2 = 1.25/[2^(n-2)] + (n-2)/2
所以an - an-1 = 1.25/[2^(n-2)] + (n-2)/2
按照规律可知,an-1 - an-2 = 1.25/[2^(n-3)] + (n-3)/2
……
a2 - a1 = 1.25
将上面n-1的式子相加就可以得到an的通项公式了.

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