an=1/[(2n+1)(2n+3)],求Sn.也就是：an等于1除以（2n+1）(2n+3)的商,求Sn.我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/（2n+1）-1/(2n+3)],接下来怎么办?

已知an=5n(n+1)(n+2)(n+3),求数列{an}的前n项和Sn 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于? 数列{an}中,a1=2,an+1=4an-3n+1,求证{an-n}是等比数列 4an中n为下标an+1中n+1为下标an-n中an的n为下标 已知a1=2 a(n+1)=2an+2^n+3^n 求an an=(3n-2)(1/2)^(n-1),求{an}前n项和 数列an,an=(2n-1)+1/【3n(n+1)】,求Sn 求数列an=n(n+1) 的前n项和 到 an=n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3（裂项） 已知数列{an}满足a1=1,且an=1/3a(n-1)+(1/3)^n (n≥2,且n∈N+),则数列{an}的通项公式为A.an=3^n/(n+2) B.an=(n+2)/3^n C.an=n+2 D.an=(n+2)3^n an=(2n-1)3^n,Sn=? 已知an=(2n+1)*3^n,求Sn f(n+1)>f(n),f(f(n))=3n.n属于正整数.令an=f(3*n次方),证明n/4n+2 已知数列{an},其中a1=1,a（n+1）=3^(2n-1)*an(n∈N),数列{bn}的前n项和Sn=log3(an/9^n)(n∈N)求an bn An=C(1,n)a1+C(2,n)a2+…C(n,n)an,若an=1+2+3+……+n(n∈N),试用n表示An. 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 已知数列｛an｝中,a(n+1)=an+2^n,a1=3,求an 裂项相消法求和（1）an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题 裂项相消法求和（1）an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题