设a1=5 an+1=2an+3^n（n≥1） 求an通项,我是新手啦.

设数列{an}中,a1=2,an+1=an+n+1,则通项an=? 已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n)an,求bn的最小值? 设数列﹛an﹜中,a1+4,an=3a(n-1)+2n-1,求通项an 设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an 设数列an满足a1=2,a(n+1)=3an+2^(n-1),求an2,设数列an满足a1=2,a(n+1)=3an+2n,求an 设a1,a2,...,an都是正数,证明不等式(a1+a2+...+an)[1/(a1)+1/(a2)+...+1/(an)]>=n^2 设数列｛an｝中,a1=1且(2n+1)an=(2n-3)a(n-1),(n大于等于2）,求{an},sn 设数列an=n^2+λn,a1 设数列﹛an﹜的前n项和为Sn,已知a1=8,an+1=Sn+3^(n+1)+5,n∈N*.设bn=an-2*3^n,证明﹛bn﹜是 等比数列 设a1=5 an+1=2an+3^n（n≥1） 求an通项,我是新手啦. 已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n）an,求bn最小值急啊 设数列{an},a1=3,a(n+1)=3an -2 (1)求证:数列{an-1}为等比数列 数列an满足a1=1/3,Sn=n(2n-1)an,求an 数列{an},a1=3,an*a(n+1)=(1/2)^n,求an a1=5,n大于等于2时,an-an-1=6n,求an {an}满足a1=2,an+1=3an+3^(n+1)-2^n（n∈正整数）,设bn=(an-2^n)/3^n,证明bn为等差数列,并求an的通项公式