an=(2^n+1/(2^n))^2求Sn

等差数列题目S(n+1)/Sn=(n+2)/n,求An 数列{an}前n项和为sn,若sn/n=3s(n-1)/n-1(n≥2),a1=3,求an an=2 S（n-1）=an 求an 两个等差数列{an},(bn}前n项和分别为Sn,S'n,若Sn/S'n=(2n+3)/(3n-1)求a9/b9 数列a(n)=n (n+1)(n+2)(n+3),求S(n) 已知数列通项公式an=n^2-n,求前n项和S an=n*2^n,求Sn an=2^n+n,求Sn 等差数列问题,疑惑,等差数列an bn的前n项和是S T S/T=2n/(3n+1)求an/bn怎么求, 已知等差数列an中,a1=1,前n项和Sn,若S(n+1)/Sn=(4n+2)/(n+1),求an 已知等差数列an中,a1=1,前n项和Sn,若S(n+1)/Sn=(4n+2)/(n+1),求an 已知a1=3且an=S(n-1)+2n,求an及snS(n-1）中n-1是下标 数列an前n项和为sn,a1=1,2s(n+1）-sn=2.n∈n*.求an的通项公式 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 数列{an}中,a1=1,an+1/an=n/n+2,求an 已知数列{an}的前n项和满足a1=1/2,an=-Sn*S(n-1),(n大于或等于2）,求an,Sn 数列an中前n项和Sn,a1=4,n≥2时,an=[√Sn+√S(n-1)]/2,求an 数列｛an｝中,a1=1,Sn=[S(n-1)]/[1+2S(n-1)](n≥2)求an