（1）由于fn(1)=a1+a2+a3+...+an = n^2,又fn(-1)=-a1+a2-a3+.+an = n,兩式相加,有2*(a2+a4+a6+...an) = n^2+n; 兩式相減有2* (a1+a3+a5+...+a(n-1)) = n^2-n,由等差數(shù)列定義,a2=a1+d,a4=a3+d...,上面兩式相減有2 * (d*n/2) = n^2+n-(n^2-n) = 2n,有d=2.根據(jù)等差數(shù)列求和公式 a1+a2+...+an = [2*a1+d(n-1)]*n/2 = n^2,有a1=1.
既通項(xiàng)公式為an=a1+d(n-1)=1+2*(n-1)=2n-1
（2）fn(1/2)= 1/2 * 1 + 1/4 * 3 + 1/8 * 5 + ...+ 1/(2^n) * (2n-1)
由于每項(xiàng)均為正,故n增大時(shí)fn(1/2)值必然增大,故fn(1/2)>= f3(1/2)= 15/8 > 5/4.
易知f5(1/2)=83/32.而第7項(xiàng) :第6項(xiàng) = (13/128) :(11/64) = 13/22.且之后任意相鄰兩項(xiàng)之比均不大于13/22,并趨近于1/2.
故fn(1/2) = f5(1/2) + 11/64 + 13/128 + 15/256 + .< f5(1/2) + 11/64 + 11/64*(13/22) + 11/64*[(13/22)^2] + .
后面是一個(gè)無窮等比數(shù)列,Sn = (11/64 * 1) / [1- (13/22)] = 121/288.
于是fn(1/2) < f5(1/2) + 121/288 = 83/32 + 121/288 = 3.0138888...< 3.0139
但是在計(jì)算第8項(xiàng)的時(shí)候,本應(yīng)為15/256,但是我們即為 11/64 * 13/22 * 13/22 = 169/2816,我們將差額予以扣除,得fn(1/2) < 2.9959,證畢