6Sn=an平方+3an
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(1)由2S(n+1)+2S(n)=3a(n+1)^2可得2S(n)+2S(n-1)=3a(n)^2两式相减得2a(n+1)+2a(n)=3[a(n+1)^2-a(n)^2]由此可得a(n+1)=-a
n=1时,S1=a1=1+1=2n≥2时,Sn=n^2+1S(n-1)=(n-1)^2+1an=Sn-S(n-1)=n^2+1-(n-1)^2-1=2n-1n=1时,a1=2-1=1,与a1=2矛盾.
当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所
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
易得an=2n+1,Sn=n(n+2);bn=1/(an*an-1)=1/(4n(n+1))=1/4*(1/n-1/(n+1)),Tn=1/4*(1-1/(n+1))
把an用sn-s(n-1)替代计算
n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-(1+3a(n-1))=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n
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
1楼貌似错了!(a1^2-3a1=6a1与An^2+3An=6Sn矛盾)An^2+3An=6SnA(n+1)^2+3A(n+1)=6S(n+1)后减前得A(n+1)^2+3A(n+1)-An^2-3A
An=3S(n-1).用原式减去,得A(n+1)-An=3An.A(n+1)=4An.则An为等比数列.
An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(
4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(
再答:满意采纳,不懂追问,谢谢
An=2n+1字数限制,详见评论
sn=2n^2-3nan=Sn-S(n-1)=2n^2-3n-[2(n-1)^2-3(n-1)]=4n-5
(1)6a1=a1^2+3a1+2解得a1=1或2(2)6sn=an^2+3an+26s(n-1)=a(n-1)^2+3a(n-1)+2两式想减得6an=an^2-a(n-1)^2+3an-3a(n-
6Sn=an^2+3an+3,6S(n-1)=[a(n-1)]^2+3a(n-1)+3相差:6an=an^2+3an-[a(n-1)]^2+3a(n-1)(注意:Sn-S(n-1)=an)整理得:(a
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