sn=n2则an等于
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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
你说的应该是平方和的数列吧.解法如下:a(n)=n^2=n(n+1)-n,n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3,n=[n(n+1)-(n-1)n]/2,a(n)=[n(
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
n=1时,a1=S1=5,a3=S3-S2=8,a5=S5-S4=12,∴a1+a3+a5=5+8+12=25.故答案为:25.
∵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
采纳帮你答再问:确定?再问:答案呢。。再问:骗子。。
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+
a1=S1=1+1=2,an=Sn-Sn-1=(n2+1)-[(n-1)2+1]=2n-1,当n=1时,2n-1=1≠a1,∴an=2,n=12n−1,n≥2.答案:an=2,n=12n−1,n≥2.
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
Sn=1*2+2*2^2+3*2^3+4*2^4+……+n*2^n给此式左右乘以2得:2Sn=1*2^2+2*2^3+3*2^4+4*2^5+……+(n-1)*2^n+n*2^(n+1)第一个式子减第
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
41a1+a2+a3+a4+a5=s5
Sn=n^2+4nS(n-1)=(n-1)^2+4(n-1)=n^2+2n-3An=S(n)-S(n-1)=2n+3
Sn=an^2a1=a1^2a1=1或a1=0S2=a2^21+a2=a2^2(a2-1/2)^2=5/4a2=1/2+√5/2或a2=1/2-√5/2Sn=an^2Sn-1=an-1^2an=Sn-
Sn=n^2+5n,a6+a7+a8+...+a16=S16-S5=336-50=286.
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
∵数列{an}的前n项和Sn=n2-9n,∴a5=S5-S4=(25-45)-(16-36)=0.故答案为:0.
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
根据数列前n项和的性质,得n≥2时,an=Sn-Sn-1=(n2-n+1)-[(n-1)2-(n-1)+1]=2n-2,当n=1时,S1=a1=1,故an=1,n=12n−2,n≥2据通项公式得|a1