设a1=√2,an+1=√2an
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a2=a1+1/a1=2+1/2=5/2a3=a2+1/a2=2/5+5/2=29/10数学归纳法证明n=1时a1=2>根号3,成立假设n=k时成立A(k)>√(2k+1)令A(k)^2=(2k+1)
A(n+1)=(1+1/n)An+(n+1)/2^nA(n+1)=(n+1)/n×An+(n+1)/2^n两边除n+1A(n+1)/(n+1)=An/n+1/2^nB(n+1)=Bn+1/2^nBn=
(1)an 的通项公式an=(-1)^(n-1)*a+3/2*n-111/4+ (105/4)*((-1)^(n-1))
a(n+1)=a(n)+n+1,a(n)=a(n-1)+(n-1)+1,...a(2)=a(1)+1+1,等号两边求和.有,a(n+1)+a(n)+...+a(2)=a(n)+...+a(2)+a(1
如果您满意我的回答,手机提问的朋友在客户端右上角评价点【满意】即可!再问:你的图看不到再答:n=1时,2a1-a1=a1=S1S1=a1²a1²-a1=0a1(a1-1)=0a1=
an1里的n1是下标吗再问:嗯再答:等一下哈,我在写漂亮点,然后拍下来给你看再答:再问:2+3+4+5+...+n是怎么等于下面那个式子的。再问:2+3+4+5+...+n是怎么等于下面那个式子的。再
(1)根据题意,有An=(An-An-1)+(An-1-An-2)+…+(A2-A1)+A1=3-2^(2n-3)+3-2^(2n-5)+…+(3-2^3)+2再用分组求和法:=3n-【2^(2n-3
a1+a2+a3+…+an=n²an,①以n+1代n,得a1+a2+a3+…+a=(n+1)²a,②②-①,a=(n+1)^2*a-n^2*an,∴a=nan/(n+2),③1)n
(1)∵a(n+1)=2an+1∴a(n+1)+1=2(an+1)∴[a(n+1)+1]/(an+1)=2∵bn=an+1a1=1,b1=2,∴bn是等比数列(2)∵bn公比是2∴bn=2^n∵bn=
S2-S1=(an+1-a1)+(an+2-a2)+...+(a2n-an)=nd*n=d*n^2S3-S2=(a2n+1-a1)+(a2n+2-a2)+...+(a3n-a2n)=nd*n=d*n^
n=1时,a2-a1=3;n=2时,a3-a2=3+d;n=3时,a4-a3=3+2d;...n=n时,a(n+1)-an=3+(n-1)d;左右相加,得:a(n+1)-a1=3n+n*(n-1)d/
a(n+1)-an=3*2^(2n-1)an-a(n-1)=3*2^(2n-3)...a3-a2=3*2^3a2-a1=3*2^1相加an-a1=3[2^1+2^3+2^5+2^7+...+2^(2n
由a(n+1)=an+In(1+1/n)得:an-a(n-1)=ln[1+1/(n-1)]a(n-1)-a(n-2)=ln[1+1/(n-2)]……a2-a1=ln(1+1/1)把上面一串式子加起来,
(1)a1+3a2+…+3^(n-2)an-1=(n-1)/3a1+3a2+…+3^(n-1)an=(n-1)/3+3^(n-1)an=n/3an=(1/3)^n.(2)bn=n/an=n3^nSn=
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
方法一:A(n+1)-1=3An-3=3(An-1),且A1-1=2,所以数列{An-1}为公比为3,首项为2的等比数列方法二:设A(n+1)+k=3(an+k),即A(n+1)=3An+2k,则2k
1、a(n+1)/an=(n+2)/(n+1)a(n+1)/(n+2)=an/(n+1)设cn=an/(n+1)则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1即c(n+1)=c
(1)由a1+a2+a3=33得:3a2=33故a2=11又由an-2+an-1+an=153【估计你这里少打了个n】得3an-1=153故an-1=51而a1+a2+...+an=n(a1+an)/
a2=a1+1/4=a+1/4.a3=(1/2)a2=a/2+1/8.b1=a1-1/4≠0.b=a-1/4=(1/2)a-1/4=(1/2)[a+1/4]-1/4=(1/2)[a-1/4]=(1/2