已知SN=n方减2n 求An

Sn=2n^2-3nSn-1=2(n-1)^2-3(n-1)an=Sn-Sn-1=-2n+2an-1=-2(n-1)+2an-an-1=-2所以an是等差数列其实数列{an}是等差数列的充要条件就是前

由题得：an=1/2(1/n-1/(n+1);所以：a1=1/2(1-1/2);a2=1/2(1/2-1/3);a3=1/2(1/3-1/4);.an=1/2(1/n-1/(n+1);sn=a1+a2

Sn=25n-n^2a1=S1=25*1-1^2=24n≥2时an=Sn-S(n-1)=25n-n^2-[25(n-1)-(n-1)^2]=26-2n令an=0得n=13所以当n≤13时Tn=|a1|

当n=1时,A1=S1=2*1^2=2；当n>1时：Sn=2*n^2S(n-1)=2*(n-1)^2=2(n^2-2n+1)=2*n^2-4n+2所以An=Sn-S(n-1)=(2*n^2)-(2*n

Sn=2^n-1---------(1)当n=1时,a1=1S(n-1)=2^(n-1)-1-------(2)(1)-(2)Sn-S(n-1)=2^n-2^(n-1)an=2^(n-1)a1+a3+

sn=2n^2-n+2s(n-1)=2(n-1)^2-(n-1)+2两式相减an=4n-3

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

sn=n^2ans(n-1)=(n-1)^2*a(n-1)sn-s(n-1)=n^2an-(n-1)^2*a(n-1)=an(n^2-1)an=(n-1)^2a(n-1)(n+1)an=(n-1)a(

n=1时,a1=S1=4×1²+2×1=6n≥2时,an=Sn-S(n-1)=4n²+2n-[4(n-1)²+2(n-1)]=8n-2n=1时,a1=8×1-2=6,同样

Sn=3＋2^nSn-1=3＋2^n-1an=sn-sn-1=3＋2^n-3-2^(n-1)=2^n-2^(n-1)=2*2^(n-1)-2^(n-1)=2^(n-1)

A(n+1)=S(n+1)-Sn=2(n+1)^2+3(n+1)+2-2n^2-3n-2=2n^2+4n+2+3n+3-2n^2-3n=4n+5An=5+4(n-1)

由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所

【方法1：强行展开a(n)表达式】1+2+……+n=n(n+1)/21^2+2^2+……+n^2=n(n+1)(2n+1)/61^3+2^3+……+n^3=n^2(n+1)^2/41^4+2^4+……

Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…

sn=a1+a2+a3+.+an=（1^2+2^2+3^2+.+n^2）-（1+2+3+...+n）+2n=n（n+1）（n+2）/6-n（1+n）/2+2n再问：三次方？这是什么数列？再答：an=n

Sn=n^2+2n-1,S1=1^2+2-1=2an=Sn-S(n-1)=n^2+2n-1-[(n-1)^2+2(n-1)-1]=2n+1a1=3所以a1≠S1当n=1时an=2当n>1时an=2n+

n≥2时,a(n)=S(n)-S(n-1)=(2n²+3n)-[2(n-1)²+3(n-1)]=4n+1当n=1时,a1=S1=2×1+3×1=5,也适合上面式子∴a(n)=4n+

错位相减Sn=1*3+3*3^2+5*3^3+.+(2n-3)*3^(n-1)+(2n-1)*3^n(1)同乘以33Sn=1*3^2+3*3^3+.+(2n-3)*3^n+(2n-1)*3^(n+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