sn=3n平方 2n 1 sn=3n次方 b 求通项公式

Sn=3n^2-2nan=Sn-S(n-1)=3n^2-2n-3(n-1)^2+2(n-1)=6n-5a1=1,S1=1an=6n-5

n(n+1)(2n+1)/6

我就说第二问吧.若{an}中存在三项,它们可以构成等差数列,则有2an=(an-1)+(an+1)即2*（3*2^n-3）=3*2^（n+1）-3+3*2^（n-1）-3,3*2^(n+1)-6=3*

a1=S1=3+2=5Sn=3n²+2n①S(n-1)=3(n-1)²+2(n-1)②①-②得an=6n-1d=an-a(n-1)=6n-1-6(n-1)+1=6

解Sn=2n²-3nS(n-1)=2(n-1)²-3(n-1)(n≥2)an=Sn-S(n-1)=2n²-3n-2(n-1)²+3(n-1)=4n-5当n=1时

Sn=3n的平方+2nSn-1=3(n-1)^2+2(n-1)An=Sn-Sn-1=3n^2+2n-3(n-1)^2-2(n-1)=3n^2+2n-3n^2+6n-3-2n+2=6n-1

an=(2n+1)*3^na1=3*3^1a2=5*3^2a3=7*3^3.an=(2n+1)*3^nSn=3*3^1+5*3^2+7*3^3+.(2n+1)*3^n3Sn=3*3^2+5*3^3+7

(2n-1)²=4n²-4n+1所以Sn=4*(1²+2²+……+n²)-4(1+2+……+n)+1*n=4*n(n+1)(2n+1)/6-4*n(n

（1）证明：由Sn=2an-3n，得Sn-1=2an-1-3（n-1）（n≥2），则有an=2an-2an-1-3an+3=2（an-1+3）（n≥2），∵a1=S1=2a1-3，∴a1=3，∴a1+

sn=2n^2-3nS(n-1)=2(n-1)^2-3(n-1)两式相减得an=2n-2-3=2n-5所以是等差数列啊.但Sn不是了

Sn=(3n+1)/2-(n/2)an当n=1时,a1=4/3=1+1/3=1+1/[1*(1+2)]当n=2时,a2=13/12=1+1/[2*(1+2+3)当n=3时,a3=31/30=1+1/[

再答：满意采纳，不懂追问，谢谢

1.k=0Sn=2n^2-3nS(n-1)=2(n-1)^2-3(n-1)an=Sn-S(n-1)=4n-5(n=1也成立)2.k≠0Sn=2n^2-3nS(n-1)=2(n-1)^2-3(n-1)a

 再问： 再问：那个划横线的答案是不是错了再答：我觉得是

1.Sn=2n^2+3nS(n-1)=2(n-1)^2+3(n-1)=2n^2-n-1Sn-S(n-1)=an=4n+12.Sn=2*3^n-1S(n-1)=2*3^(n-1)-1Sn-S(n-1)=

sn=2n^2-3nan=Sn-S(n-1)=2n^2-3n-[2(n-1)^2-3(n-1)]=4n-5

Sn=2+5n+8n^2+…+(3n-1)n^n-1nSn=2n+5n^2+…+(3n-4)n^(n-1)+(3n-1)n^nSn-nSn=2+3n+3n^2+…+3n^(n-1)-(3n-1)n^n

(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1

Sn=1^2-2^2+3^2-4^2+5^2-6^2+...+(-1)^(n-1)*n^2n为奇数时Sn=1^2+(-2^2+3^2)+(-4^2+5^2)+...+(-(n-1)^2+n^2)=1+

看不懂啊是Sn=2n^2-(3n+1)还是Sn=(2n)^2-(3n+1)?题目容易令n=1求出a1=-2Sn-1=2(n-1)^2-3(3(n-1)+1)an=Sn-Sn-1=2(2n-1)-3=4