数列an中a1等于1对所有的n≥2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/08 07:04:04
由:a1+2a2+2^2a3+…+2^(n-1)an=8-5n--------------------------------①知:a1+2a2+2^2a3+…+2^(n-2)a(n-1)=8-5(n
a1a2=4,a1a2a3=9,所以a3=9/4,a1a2a3a4=16,a1a2a3a4a5=25,所以a5=25/16a3+a5=9/4+25/16=61/16
∵an=an-1+1/n(n+1)∴an-an-1=1/n-1/(n+1)an-1-an-2=1/(n-1)-1/n………a2-a1=1-1/2上述各式相加得:an-a1=1-1/(n+1)=n/(n
由题意a1a2…an=n2,故a1a2…an-1=(n-1)2,两式相除得:an=n2(n−1)2 (n≥2),所以a3=94,a5=2516,即a3+a5=6116故选B.
由an+1=an+2n可以列出以下各式a10=a9+2x9a9=a8+2x8a8=a7+2x7..a3=a2+2x2a2=a1+2x1以上各式相加可得a10=a1+1x2+2x2+.+9x2a10=9
1.变形即为a(n+1)-(n+1)=4(an-n),所以(an-n)是首项为1,公比为4的等比数列.2.令an-n=bn,则Sbn=(4^n-1)/(4-1),即San-1-2-…-n=(4^n-1
应该是A(n+1)=An+2n吧~~~=>a(n+1)-an=2n所以an-a(n-1)=2(n-1)a(n-1)-a(n-2)=2(n-2)...a2-a1=2*1把左边加起来,右边加起来得到an-
当n≥2时,由a1•a2•a3…an=n2①,得a1•a2•a3…an-1=(n-1)2②,①②得an=n2(n−1)2,又a1=1,∴an=1(n=1)n2(n−1)2(n≥2),故答案为:1(n=
an+1=(1-1/n+1)an则an+1=(n/n+1)an则an+1=(n/n+1)an=(n/n+1)*(n-1/n)an-1=...=n/n+1*(n-1/n)*..1/2*a1=1/n+1所
a1=1a1*a2…*an=n^2得a2=4得an=n^2/(n-1)^2得a3=9/4a5=25/16所以a3+a5=61/16
当n≥2时,a1•a2•a3••an=n2.当n≥3时,a1•a2•a3••an-1=(n-1)2.两式相除an=(nn−1)2,∴a3=94,a5=2516.∴a3+a5=6116.故选A
由已知an与1的等差中项等于Sn与1的等比中项得(an+1)/2=√SnSn=(an+1)²/4n=1时,S1=a1=(a1+1)²/4,整理,得(a1-1)²=0a1=
Sn=a1+a2+...+an=2^n-11.n=1时,a1=S1=2-1=12.n>=2时,an=Sn-S(n-1)=2^n-2^(n-1)=2^(n-1),a1=1符合故an=2^(n-1)数列是
(1)用数学归纳法.A(n+1)=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2(2)因为an>=n+2,所以an-n>=2A(n+1)=An(
由a1=1,a1·a2=2^2=4,则a2=4,同理,a3=9/4,a4=16/9,a5=25/16.这是很死的方法了.灵活地点就这样了,a1·a2·a3·a4=16,a1·a2·a3·a4·a5=2
a1*a2*a3*.*an=n^2;a1*a2*a3*.*a(n-1)=(n-1)^2两式相除,得到an=(n/n-1)^2则a3=1.25;a5=1.5625相加得到a3+a5=2.8125
an=Sn-S(n-1)=2^n-1-[2^(n-1)-1]=2^(n-1)an^2=4^(n-1)a1^2=1a1^2+a2^2+...+an^2=(1-4^n)/(1-4)=(4^n-1)/3
其实这是数列,a1*a2*..*a(n-1)=(n-1)的平方所以an=n的平方/(n-1)的平方.所以a3=9/4,a5=25/16.所以a3+a5=61/16.
A3=(A1*A2*A3)/(A1*A2)=3²/2²=9/4A5=(A1*A2*A3*A4*A5)/(A1*A2*A3*A4)=5²/4²=25/16A3+A
∵an与1的差数中项等于√Sn∴a[n]+1=2√S[n]两边平方(a[n]+1)²=4S[n]①∴(a[n-1]+1)²=4S[n-1]②两式相减(a[n]+1)²-(