已知数列满足2an,n=11,12
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an=1+2+3+…+n=[n(n+1)]/2则:1/(an)=2/[n(n+1)]=2[(1/n)-1/(n+1)],所以:M=1/(a1)+1/(a2)+1/(a3)+…+1/(an)=2[1/1
an/a(n-1)=(n+1)/(n-1)(n>=2)a(n-1)/a(n-2)=n(n-2)...a2/a1=3/1上式全部相乘an/a1=(n+1)!/2(n-1)!=n(n+1)/2,an=n(
(1)bn=a(2n+1)+4n-2b(n+1)=a(2n+3)+4(n+1)-2=a(2n+2+1)+4n+2=a(2n+2)-2(2n+2)+4n+2=a(2n+1+1)-2(2n+2)+4n+2
a2-a1=2,a3-a2=4,…an+1-an=2n,这n个式子相加,就有an+1=100+n(n+1),即an=n(n-1)+100=n2-n+100,∴ann=n+100n-1≥2n•100n-
(1)由已知a2=2a1+2,a3=2a2+3=4a1+7,若{an}是等差数列,则2a2=a1+a3,即4a1+4=5a1+7,得a1=-3,a2=-4,故d=-1. &nbs
你把这个数列看成俩部分a(n1)=2a(n1-1)a(n2)=2n+2an=(an1)+(an2)算算看
由题意,Sn=n^2,则a1=1,S(n-1)=(n-1)^2=n^2-2n+1,n>=2an=Sn-S(n-1)=n^2-n^2+2n-1=2n-1,n>=2由于当n=1时,2n-1=1=a1所以,
(1)证b1=a2-a1=1,当n≥2时,bn=an+1−an=an−1+an2−an=−12(an−an−1)=−12bn−1,所以{bn}是以1为首项,−12为公比的等比数列.(2)解由(1)知b
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-
a(n+1)-2an=3.5^n,则a2-2a1=3.5^1a3-2a2=3.5^2.a(n+1)-2an=3.5^n以上式子相加,得a(n+1)-a1-Sn=3.5+3.5^2+...+3.5^n=
(1)dn满足dn=[3+(-1)的n次方]/2易知,dn=1n是奇数dn=2n是偶数又由an=d1+d2+d3+...d2n,得d1+d2=d3+d4=.,所以通项公式an=3n且b2,b4为方程x
你应该是抄错题了吧--A(n+1)=2An+2^n等式两边同时除以2^(n+1)有A(n+1)/2^n+1=An/2^n+1/2设Bn=An/2^n则B(n+1)=Bn+0.5Bn是等差数列即An/2
1.a_(1)=1,a_(n+1)=2a_(n)+2^(n)----------------1b_(n)=a_(n)/2^(n)将式子1左右两边同时除以2^(n+1),则:b_(n+1)=b_(n)+
a1=2>0假设当n=k(k∈N+)时,ak>0,则a(k+1)=3√ak>0k为任意正整数,因此对于任意正整数n,an恒>0,数列各项均为正.a(n+1)=3√anlog3[a(n+1)]=log3
将已知等式取倒数,得1/an=[3a(n-1)+1]/a(n-1)=1/a(n-1)+3,所以,{1/an}是首项为1/a1=1,公差为3的等差数列,因此1/an=1+3(n-1)=3n-2,所以an
1)累加法a1=2a2-a1=1/(1*2)a3-a2=1/(2*3)a4-a3=1/(3*4).an-a(n-1)=1/[(n-1)n]相加得an=2+(1-1/2)+(1/2-1/3)+(1/3-
∵s[n]=n^2a[n]∴s[n+1]=(n+1)^2a[n+1]将上述两式相减,得:a[n+1]=(n+1)^2a[n+1]-n^2a[n](n^2+2n)a[n+1]=n^2a[n]即:a[n+
/>n≥2时,Sn=n²×anS(n-1)=(n-1)²×a(n-1)an=Sn-S(n-1)=n²×an-(n-1)²×a(n-1)(n²-1)an