已知Sn=3n的平方 b

Sn=2n²+3n+1①则Sn-1=2(n-1)²+3（n-1)+1②(n＞1）①式减②式得an=4n+1（n＞1）当n=1a1=s1=2+3+1=6不符合an的式子故an=6(n

因为S(n+1)-S(n)=A(n+1),根据题意有:2S(n+1)^2=2A(n+1)S(n+1)-A(n+1),将上式代入此式得:2S(n+1)^2=2[S(n+1)-S(n)]S(n+1)-S(

an+1=Sn+3^nS(n+1)=S(n)+a(n+1)=2Sn+3^nS(n+1)-3^(n+1)=2[s(n)-3^n]即b(n+1)=2b(n)bn为等比数列,公比为2b1=S1-3^1=a1

a1=s1=1/4+2/3+3=47/12An=Sn-S(n-1)=1/4n^2+2/3n+3-[1/4(n-1)^2+2/3(n-1)+3]An=1/2n+5/12n≥2A1=47/12注意AN.要

向量a=(n,Sn)b=(4,n+3)共线所以n(n+3)-4Sn=0Sn=n(n+3)/4a1=S1=1n>=2:an=Sn-S(n-1)=n(n+3)/4-(n-1)(n+2)/4=(n+1)/2

∵sn=n^2+n/2∴an=sn-s(n-1)=n^2+n/2-(n-1)^2-(n-1)/2=2n-1/2

1:Sn=2n^2-3n+1Sn-1=2n^2-7n+6an=Sn-Sn-1=4n-52:a1+a1q^3=18a1q+a1q^2=12(1+q^3)/(q+q^2)=3/2(q-2)(2q^2+q-

1.n>1,3Sn^2=An(3Sn-1)An=Sn-S(n-1)下标3Sn^2=(3Sn-1)[Sn-S(n-1)]=3Sn^2-3SnS(n-1)-Sn+S(n-1)3SnS(n-1)=S(n-1

2^(n+1)-2^n=2*2^n-2^n=2^nb*an-2^n=(b-1)Sn,b*a(n+1)-2^(n+1)=(b-1)S(n+1)两式相减(左-左=右-右)：[b*a(n+1)-2^(n+1

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

S1=a1=13(S2)^2=a2(3S1)S2=a1+a2=1+a2,代入上式得：3*(1+a2)^2=3*a2-->a2^2+a2+1=0此方程无实数解,题目问题.再问：我算出来了a2是-3/4,

答：等比数列an的前n项和公式为Sn=3^n+2bS(n+1)=3^(n+1)+2b两式相减：A(n+1)=S(n+1)-Sn=2*3^n所以：A1=2=3+2b解得：b=-1/2

Sn^2-n^2×Sn-(n^2+1)=0(Sn+1)[Sn-(n^2+1)]=0数列各项为非零实数,S1≠0,且Sn不恒为0,因此只有Sn=n^2+1n=1时,a1=S1=1+1=2n≥2时,an=

再答：满意采纳，不懂追问，谢谢

Sn=a*n^2+b*n+c,——》S(n-1)=a*(n-1)^2+b*(n-1)+c,——》an=Sn-S(n-1)=(a*n^2+b*n+c)-[a*(n-1)^2+b*(n-1)+c]=(2n

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)=

a8+a9+a10+a11+a12=S12-S7=(12^2+12+1)-(7^2+7+1)=100

an=Sn－Sn－1=3n²＋n－3﹙n－1﹚²－﹙n－1﹚=6n－2再问：a1=20,an=54,Sn=999，求d及n,d=2,n=15.An-2=-14，求a5及Sn谢谢哈。

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

(n+1)³-n³=3n²+3n+1n³-(n-1)³=3(n-1)²+3(n-1)+1……2³-1³=3×1²