已知数列an中an>0对于任意m,n正整数都有an*am=2^m n
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/29 11:32:08
计算出a1=1,a2=根号2-1,a3=根号3-2猜想an=根号n-根号(n-1),Sn=根号n用数学归纳法证明n=1时2a1=a1+1/a1,a1=1成立假设n=k成立,则n=k+1时2√k+2a(
令n=2,m=1:(a2)²-(a1)²=a1a3;所以a3=-1;令n=n>2,m=2:(an)²-(a2)²=(an-2)(an+2),(an)²
因为an-2/an=2n所以:(an)^2-2nan-2=0根据万能公式:an=n-√(n^2+2),an=n+√(n^2+2)>0又因an<0所以:an=n-√(n^2+2),假设m>n>0那么am
∵数列{an}中,an=2n−1(n为正奇数)2n−1(n为正偶数),∴a9=29-1=28=256.S9=21-1+(2×2-1)+23-1+(2×4-1)+25-1+(2×6-1)+27-1+(2
∵an+1=an+n,∴an+1-an=n,∴an=a1+(a2-a1)+…+(an-an-1)=1+1+2+…+(n-1)=1+n(n−1)2∴a100=1+100×992=4951.
由a(n+2)=5a(n+1)-6a(n)知a(n+2)-2a(n+1)=3[a(n+1)-2a(n)]即b(n+1)=3bn则{bn}为等比数列易求得{bn}通项公式bn=2*3^n由bn=a(n+
∵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
2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1
这种方法看似麻烦,实际很简单~由上式可得:an+2=(1+an+1)/(1-an+1)代入an+1的表达式,化简得到an+2=-1/an同样的方法代入an+2可以得到an+3=(an-1)/(an+1
点(an,an-1)在直线x-y-6=0上有an-a(n-1)-6=0an-a(n-1)=6故{an}以3为首项,6为公差的等差数列故an=3+(n-1)6=6n-3故a3=15,a5=27,a7=3
1.由已知:ap+aq=a(p+q)可令p=n,q=1,则有:an+a1=a(n+1)a(n+1)-an=a1=2故,数列{an}是首项为2,公差也为2的等差数列,其通项公式为:an=2+2*(n-1
解题思路:构造数列解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph
∵a1=1,an+1=an+n,∴a2-a1=1,a3-a2=2,…,a100-a99=99,∴a100=a1+(a2-a1)+(a3-a2)+…+(a100-a99)=1+1+2+…+99=4951
A(n+1)-A(n)=2n+1-λ>0λλ<3
由已知条件列式:(an+2)/2=√2Sn整理,得8Sn=(an+2)²令n=1S1=a1代入,整理,得(a1-2)²=0a1-2=0a1=2令n=2S2=2+a2代入,整理,得a
an-a(n-1)(n>2)=n^2+λn-(n-1)^2-λ(n-1)=n^2-(n-1)^2+λ=2n-1-λ数列an是递增数列2n-1+λ>0λ>1-2nn>2λ>-3
∵(an)²≤an-a(n+1),得a(n+1)≤an-(an)²∵在数列{an}中an>0,∴a(n+1)>0,∴an-(an)²>0,∴0<an<1故数列{an}中的
对于任意的n≥2(n为自然数),由3Sn-4,an,2-3/2Sn-1(n-1为下标)总成等差数列,得2an=3Sn-4+2-3/2Sn-1即2an=3/2Sn+3/2(Sn-Sn-1)-2=3/2S
因为数列{an}中,a1=1,且对于任意的正整数m,n都有am+n=aman+am+an,∴an+1=ana1+an+a1=2an+1;∴an+1+1=2(an+1);∴an+1+1an+1=2;故数
∵an=nn2+156=1n+156n≤1439∵1n+156n≤1439当且仅当n=239时取等,又由n∈N+,故数列{an}的最大项可能为第12项或第13项又∵当n=12时,a12=12122+1