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数列满足a1
数列
an,
满足a1
=1,a2=2,a3=3,a4=4,a5=5,当n>=5时an+1=a1a2……an-1,问...
答:
a(n+1)=
a1
a2……a(n-1)a(n+2)=a1a2……a(n-1)an=a(n+1)an a6=a1a2a3a4=24 a7=a1a2a3a4a5=a5a6=5*24 a8=a7a6=a6a5a6=a5a6^2=5*24^2 a9=a8a7=a5a6^2*a5a6=a5^2*a6^3=5^2*24^3 a(n+2)=a(n+1)an=ana(n+1)=a(n-1)an^2 =a(n-1)[a(n-1)a(n...
已知正项等比
数列满足a1
a3等于16,a5等于32,求数列(an)的通项公式_百度...
答:
an是正项等比
数列
,因此,an>0,可设an=
a1
q^n 所以,a1a3=a1a1q^2=16 a5=a1q^4=32 得:(a1q)^2=16 a1q^4=32 因为,an>0,且是正项数列,所以:a1q=4 a1q^4=32 q^3=8 q=2 a1=2 因此:an=2×2^(n-1)=2^n
已知
数列
an
满足a1
等于1,
答:
即1/an<1/3^(n-1)1/
a1
+1/a2+...+1/an<1/3^0+1/3^1+1/3^2+...+1/3^(n-1)=[1-(1/3)^n]/(1-1/3)=3/2*[1-(1/3)^n]<3/2 所以1/a1+1/a2+...+1/an<3/2
已知
数列
{an}
满足a1
=1/3,a2=7/9,an+2=4/3an+1-1/3an (1)求{an}的通...
答:
{a(n+1)-a(n)}是首项为a(2)-a(1)=7/9 - 1/3 = 4/9,公比为(1/3)的等比
数列
.a(n+1)-a(n) = (4/9)(1/3)^(n-1) = 4/3^(n+1),a(n+1)3^(n+1) = 3a(n)3^(n) + 4,2+a(n+1)3^(n+1) = 3[2 + a(n)3^n]{2+a(n)3^(n)}是首项为2+3a(...
已知
数列
An
满足a1
=1,an=a1+2a2+...+(n-1)an-1,求通项公式。要有详细步骤...
答:
如果是a2=
a1
=1,那么叠乘的工作进行到a2即可,n=1时单独写通项公式.最后通项公式的形式分n≥2和n=1两种情况,写成分段函数的形式就可以了。
数列
[a]
满足a1
=1数列[2^n/an]是公差为1的等差数列 (1)求数列[an]的通...
答:
解:1、2/
a1
=2/1=2 2ⁿ/an=2+1×(n-1)=n+1 an=2ⁿ/(n+1)n=1时,a1=2/(1+1)=1,同样
满足
。
数列
{an}的通项公式为an=2ⁿ/(n+1)。2.bn=n(n+1)an=n(n+1)2ⁿ/(n+1)=n×2ⁿSn=b1+b2+...+bn=1×2+2×2²+3×2³...
已知
数列
{an}
满足a1
=1,an=a1+2a2+3a3+…(n-1)an-1(n>=2)则{an}的通...
答:
则a(n+1)=
a1
+2a2+3a3+…(n-1)an-1+nan ...② 两式相减②-① 得a(n+1)-an=nan (n>=3)即a(n+1)=(n+1)an 即 a4=3a3 a5=4a4 ...a(n-1)=(n-1)a(n-2)an=na(n-1)上述各式相乘得 an=n(n-1)(n-2)*...*4*3 =n(n-1)(n-2)*...*4*...
数列
{an}
满足a1
=1,设该数列的前n项和为Sn,且Sn,Sn+1,2a1成等差数列.用...
答:
解析:由题意 2Sn+1=Sn+2
a1
=Sn+2 归纳法证明 当n=1时,S1=
a1
=1
满足
式子 假设n=k时,成立即Sk=(2k-1)/2k-1 则n=k+1时,Sk+1=1/2Sk+1=(2k-1)/2k +1=(2k+1-1)/2k 即n=k+1时,等式成立 所以可以证明式子对所有n成立。
已知
数列
{an}
满足a1
=1,a(n+1)=an/(2an+1)。归纳推测an,并用数学归纳法...
答:
由递推公式可得
a1
=1 ,a2=1/3 ,a3=1/5 ,a4=1/7 ,推测 an=1/(2n-1) 。证明:(1)当 n=1 时,显然成立 ,(2)设当 n=k 时有 ak=1/(2k-1)(k>=1) ,则当 n=k+1 时有 a(k+1)=ak/(2ak+1)=[1/(2k-1)] / [2/(2k-1)+1]=[1/(2k-1)] / [(2k+1...
已知
数列满足
:
a1
=1,a(n+1)=an+1,n为奇数;2an,n为偶数,设bn=a2n-1...
答:
a1
=1 a2 =a1+1 =2 if n is odd,a(n+1) = an +1 = 2a(n-1) +1 a(n+1) +1 = 2[ (a(n-1) +1 ]a(n+1) +1 = 2^[( n-1)/2 ]. (a2 +1 )= 3.2^[( n-1)/2 ]a(n+1) = -1+3.2^[( n-1)/2 ]n is odd => n= 2m-1 a(2m) =-1+3....
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