若sn=2n^2 3n 2,则an=
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A.an=2n-1Sn=n^2Sn-1=(n-1)^2an=Sn-Sn-1=n^2-(n-1)^2=n^2-n^2+2n-1=2n-1
an=sn-s(n-1)=n^2+1-(n-1)^2-1=2n-1
∵an=Sn-Sn-1(n≥2),Sn=n2∴a8=S8-S7=64-49=15故答案为15
a6+a7+a8=S8-S5=8²+2×8+5-(5²+2×5+5)=45
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令n=7和4,由Sn=n2+2n+5得到:S7=68和S4=29,则a5+a6+a7=S7-S4=68-29=39.故答案为:39
由an+1+Sn=n2+2n①,得an+Sn-1=(n-1)2+2(n-1)(n≥2)②,①-②得,an+1=2n+1(n≥2),an=2n-1(n≥3),又a1=0,a2=3,所以an=0,n=12
(1)证明:①n=1时,a1=S1=23.②n≥2时,an=Sn-Sn-1=(25n-2n2)-[25(n-1)-2(n-1)2]=27-4n,而n=1适合该式.于是{an}为等差数列.(2)因为an
sn=a1+a2+a3+……+an=1*2^0+2*2+3*2^2+4*2^3+……+n2^(n-1)2sn=1*2+2*2^2+3*2^3+……+n*2^n两式相减得-sn=1+2+2^2+2^3+
Sn=n^2/(3n+2)Sn-1=(n-1)^2/(3n-1)an=Sn-Sn-1=(3·n^2+n-2)/(9·n^2+3n-2)所以,当n接近正无穷时liman=1/3
当n≥2时,an=Sn-Sn-1=n2-2n-1-[(n-1)2-2(n-1)-1]=2n-3,当n=1时,a1=S1=1-2-1=-2,不适合上式,∴数列{an}的通项公式an=−2,(n=1)2n
(I)a1=S1=3当n≥2时,an=Sn-Sn-1=n2+2n-[(n-1)2+2(n-1)]=2n+14,符合(II)设等比数列的公比为q,则b2=3,b4=5+7=12所以b1q=3b1q3=1
(I)当n=1时,a1=S1=4,当n≥2时,an=Sn-Sn-1=n2+2n+1-[(n-1)2+2(n-1)+1]=2n+1,又a1=4不适合上式,∴an=4,
Sn=n^2+4nS(n-1)=(n-1)^2+4(n-1)=n^2+2n-3An=S(n)-S(n-1)=2n+3
ai=1/2Sn=1/2(n2+3n-2)-anSn-1=1/2((n-1)^2+3(n-1)-2)-an-1相减2an=2n+1+an-1设参数方程求解后:an-4(n+1)+6=(1/2)^(n-
由Sn=n^2+3n得S(n-1)=(n-1)^2+3(n-1),两式相减,考虑到Sn-S(n-1)=an得an=2n-1+3=2n+4,于是得a(n-1)=2(n-1)+4,两式相减得an-a(n-
当n≥2时,an=Sn-Sn-1=2n+1∴a2=5,a3=7∴d=7-5=2a1=1+2+a=3+a∵{an}为等差数列∴a1=a2-d=3=3+a∴a=0故答案为:0
(1)当n=1时,T1=2S1-1因为T1=S1=a1,所以a1=2a1-1,求得a1=1(2)当n≥2时,Sn=Tn-Tn-1=2Sn-n2-[2Sn-1-(n-1)2]=2Sn-2Sn-1-2n+
an=sn-sn-1=n2-10n-[(n-1)2-10(n-1)]
由题意知,当n=1时,a1=s1=1+1=2,当n≥2时,an=sn-sn-1=(n2+1)-[(n-1)2+1)]=2n-1,经验证当n=1时不符合上式,∴an=22n−1n=1n≥2故答案为:an