若数列an对于一切正整数满足a1 2a2 22a3
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由an+1+an−1an+1−an+1=n可得an+1+an-1=nan+1-nan+n∴(1-n)an+1+(1+n)an=1+n∴an+1=n+1n−1an−n+1n−1=1n−1(an−1)×(
楼主英明,将a(n)=2带入已知条件,矛盾.因此,楼主的怀疑的对滴.a(n)=2肯定有问题.俺的解法如下:a(n+1)+a(n)-1=na(n+1)-na(n)+n,(n-1)a(n+1)=(n+1)
a(1)b(n)+a(2)b(n-1)+...+a(n-1)b(2)+a(n)b(1)=2^(n+1)-n-2,a(1)b(1)=2^2-1-2=1,1,a(n)=1+(n-1)=n,a(1)=1,b
由an+1=an+2n,得an+1-an=2n,∴n≥2时,a2-a1=2,a3-a2=4,…,an-an-1=2(n-1),以上各式相加,得an-a1=(n-1)(2n-2+2)2=n2-n,∵a1
(n+1)/bn=2∴bn=b1×2^(n-1)b1=a2-a1=3-1=2∴bn=2^n∴a(n+1)-an=2^n∴a2-a1=2a3-a2=2^2a4-a3=2^3……an-a(n-1)=2^(
∵an+an+1=12(n∈N*),a1=−12,S2011=a1+(a2+a3)+(a4+a5)+…+(a2010+a2011)=-12+12+…+12=−12+12×1005=502故答案为:50
a1^3+a2^3+...+an^3=sn^2a1^3+a2^3+...+[a(n+1)]^3=[s(n+1)]^2两式相减得[a(n+1)]^3=[s(n+1)]^2-sn^2[a(n+1)]^3=
(1)∵an+1=2an+1∴an+1+1=2(an+1)∴数列{an+1}是等比数列∴an+1=(a1+1)×2^(n-1)=2^n∴an=2^n-1(2)设am≤0am+1≥0∴2m-49≤02(
设A1A2=a则:由于在数列{An}中An小于0故a>0,且An+1An+2/AnAn+1>0即q>0;由题中:2AnAn+1+An+1An+2>An+2An+3得2aq^(n-1)+aq^n>aq^
因为对于任意的正整数n,恒有a2n=an+n,所以:a512=a256+256=a256+28=a128+128+256=a128+27+28=a64+26+27+28=…=a2+22+23+…+28
a(n-1)+an=4n,a(n-2)+a(n-1)=4n-4,a1=3,a2=5,an-a(n-2)=4,故a=2n+1b1+2b2+…+2^(n-1)bn=nan,b1+2b2+…+2^(n-2)
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
a1=aa2=1/a^2a3=a^4a4=1/a^8……a1*a2…*a10=1/a(1+2^2+2^4+2^6+2^8)=1/a^341
1、①A1+3A2+3^2*A3+...+3^(n-1)*An=n/3,又A1+3A2+3^2*A3+...+3^(n-)*An-1=(n-1)/3,(比已知的式子最后少写一项,即有n-1项),两式相
我高中时好像没做过这么晕头转向的数列题目……楼主好运!
(n+1)an^2+an*an+1-n(an+1)^2=0得到:((n+1)an-nan+1)(an+an+1)=0an>0,所以只有(n+1)an=nan+1所以an+1/n+1=an/n=an-1
∵(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
a100=9901a(n+1)-an=2na(n)-a(n-1)=2(n-1)a(n-1)-a(n-2)=2(n-2)………………a2-a3=4a2-a1=2累加,得:a(n)-a1=2n*½
由题意知:∵数列{1xn}为调和数列∴11xn+1−11xn=xn+1−xn=d∴{xn}是等差数列 又∵x1+x2+…+x20=200=20(x1+x20)2∴x1+x20=20又∵x1+