sn=3n平方 8n
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Sn=2*n²+1S(n-1)=2*(n-1)²+1Sn-S(n-1)=AnAn=2*n²+1-2*n²+4n-2-1An=4n-2N≥2A1=3
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
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
根据题意:S(n)=1/2+2/2²+3/2³+……++(n-1)/[2^(n-1)]+n/(2^n)(1/2)S(n)=1/2²+2/2³+3/(2^4)+…
Sn+1=2n^2+3n+1=2(n+1)^2+3(n+1)+1-4n-2-3=2(n+1)^2-(n+1)Sn=2n^2-nSn-1=2(n-1)^2-(n-1)an=Sn-Sn-1=2n^2-n-
(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
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^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
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
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
1/n^2+n=1/n(n+1)列项得1/n(n+1)=1/n-1/(n-1)然后累加
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
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。
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