等比数列和等差数列的数学问题

(1)存在常数p,q使数列{an+pn+q}成等比数列。设：a[n+1]+p(n+1)+q=2(an+pn+q),化简得：a[n+1]=2an+pn+q,对照a[n+1]=2an+n+1,得：p=1,q=1∴a[n+1]+(n+1)+1=2(an+n+1),{a[n+1]+(n+1)+1}/(an+n+1)=2,又∵a1+1+1=3所以{an+n+1}是以首项为3,公比为2的等比数列。(2)由(1)得：an+n+1=3*2^(n-1)∴an=3*2^(n-1)-n-1