1、等比数列通项bn=b1*q^(n-1),对其取对数则得到一个等差数列即:
In(bn)=In(b1)+(n-1)In(q)带入题中等式有:
(m-n)*[In(b1)+(p-1)In(q)] + (n-p)*[In(b1)+(m-1)In(q) ]+ (p-m)*[In(b1)+(n-1)In(q)] =0
对等式两边取e的对数有:
[(ap)^(m-n)]*[(am)^(n-p)]*[(an)^(p-m)]=1
2、f(n)一共有n项,则f(n+1)有n+1项.
所以f(n+1)=1/(n+2)+…………+1/(n+n)+1/(2n+1)+1/(2n+2)
则f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)=1/(2n+1)-1/(2n+2)
3、数列第[(n-1)*n/2]+1项到第n*(n+1)/2项是n
令n*(n+1)/2=2008,求得n=62.87即n=62时,项数稍小于2008,则第2008项为63