三. 已知 an=2 n^2-6n 10,则{an}的最大项是
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你是通过f=0解出ns和k0的关系么?把其他参数的数值给出来吧.再问:呃,错了,有值的n1=1.509n2=1.454n=0b=0.52ns取值1.4--1.6再答:n1=1.509;n2=1.454
(I)证明:因为an+1=2an+3an-1,所以an+1+an=3(an+an-1),所以an+1+anan+an−1=3是常数,所以数列{an+an+1}是以a1+a2=3为首项,等比为3的等比数
(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
在数列{an}中,∵2an+1=an+an+2,∴{an}为等差数列,设公差为d,由a3=a1+2d=−6S6=6a1+6×52d=−30,得a1=−10d=2.∴an=a1+(n-1)d=2n-12
(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
已知{an}满足a1=5,a2=5,an+1=an+6a(n-1)(n>=2,n∈N+),且当λ=2,或λ=-3时,数列{an+1+λan}是等比数列.1.求数列{an}的通项公式.2.设3的n次方b
(I)∵a1=20,a2=7,an+2-an=-2∴a3=18,a4=5由题意可得数列{an}奇数项、偶数项分布是以-2为公差的等差数列当n为奇数时,an=a1+(n+12−1)×(−2)=21-n当
充分性:∵an+an+1=2n+1,∴an+an+1=n+1+n,即an+1-(n+1)=-(an-n),若a1=1,则a2-(1+1)=-(a1-1)=0,∴a2=2,以此类推得到an=n,此时{a
A(n+2)=6*(n+1)*2^(n+1)-A(n+1)A(n+2)-A(n+1)=(6n+12)*2^n-A(n+1)+AnA(n+2)=(6n+12)*2^n+AnA3=37A2=11d=26A
n2=++n1先作n1=++n1,此时n1=n1+1=2+1=3,再作n2=n1=3n1=n2++先作n1=n2=3,再作n2=n2++=n2+1=3+1=4执行后n1=3,n2=4
提示哪里就是哪里出错了你调用函数fft1没有往里面传递m但是你函数里面用到m了m没定义再问:那怎么加到里面啊???再答:这函数你写的我怎么知道怎么加到里面如果不是你写的看是不是抄错了,或者把m换成n试
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
an=(n+1)(n+2)再问:有木有过程?再答:原式整理后得到an=(n+1)(an-1/n+1)试值:a2=(2+1)(6/2+1)=(2+1)(2x3/2+1)=12=3x4a3=(3+1)(1
(1)a1=3×1+6=9;a2=3×2+6=12a3=3×3+6=15b1=2×1+7=9b2=2×2+7=11b3=2×3+7=13∴c1=9;c2=11;c3=12;c4=13(2)解对于an=
(1)证明:∵an+1=Sn+1-Sn=18(an+1+2)2-18(an+2)2,∴8an+1=(an+1+2)2-(an+2)2,∴(an+1-2)2-(an+2)2=0,(an+1+an)(an
(I)∵an+1=3an-2an-1(n≥2)∴(an+1-an)=2(an-an-1)(n≥2)∵a1=2,a2=4∴a2-a1=2≠0,∴an+1-an≠0故数列{an+1-an}是公比为2的等比
an=2^n-1bn=(2n+1)(an+1)=(2n+1)*2^nTn=3*2+5*2^2+...+(2n+1)*2^n(1/2)Tn=3+5*2+7*2^2+.+(2n+1)*2^(n-1)(1/
f(0+0)=f(0)f(0)f(0)=1f(1+11)=f(1)*f(1)f(2)=4f(3)=f(1+2)=2*4=8同理f(4)=16(2)猜测f(n)=2的n次方根据f(1)=2.成立令f(n