设数列〔an〕满足a1=1，a2=5/3(3分之5），an+2=5/3an+1－2/3an,(n属
(1)a(n+2)=(5/3)*a(n+1)-(2/3)*a(n)
3a(n+2)=5a(n+1)-2*a(n)
3a(n+2)-3a(n+1)=2a(n+1)-2a(n)
3[a(n+2)-a(n+1)]=2[a(n+1)-a(n)]
[a(n+2)-a(n+1)]/[a(n+1)-a(n)]=2/3
即b(n+1)/b(n) =2/3
b(n)是等比数列
b1=a2-a1=5/3-1=2/3
b(n)是首项为2/3,公比为2/3的等比数列
b(n)=(2/3)^n
(2)an-a(n-1)=b(n-1)=(2/3)^(n-1)
a(n-1)-a(n-2)=b(n-2)=(2/3)^(n-2)
.....
a3-a2=b2=(2/3)^2
a2-a1=b1=(2/3)^1
各等式相加得
an-a1=(2/3)^1+(2/3)^2+…+(2/3)^(n-1)
an=a1+(2/3)*(1-(2/3)^(n-1))/(1-2/3)
=1+2(1-(2/3)^(n-1))
=3-2(2/3)^(n-1)
Sn=3n-2[(1-(2/3)^n)/(1-2/3)]=3n+6(2/3)^n-6
(2)
an-a(n-1)=(2/3)^(n-1)
a(n-1)-a(n-2)=(2/3)^(n-2)
.....
a3-a2=(2/3)^2
a2-a1=(2/3)^1
各等式相加 所以
an-a1=(2/3)^1+(2/3)^2+…+(2/3)^(n-1)
an=a1+(2/3)*(1-(2/3)^(n-1))/(1-2/3)
=3-2(2/3)^(n-1)
所以 Sn=6(2/3)^n+3n-6
a(n+2)=(5/3)a(n+1)-(2/3)an
a(n+2)-a(n+1) = (2/3)(a(n+1) -an)
{a(n+1) -an} 是等比数列, q=2/3
a(n+1) -an = (2/3)^(n-1).(a2-a1)
= (2/3)^n
bn = a(n+1) -an = (2/3)^n
a(n+1) - an = (2/3)^n
an -a(n-1) = (2/3)^(n-1)
an - a1 = (2/3)+(2/3)^2+....+(2/3)^(n-1)
= (2/3)(1- (2/3)^(n-1) )/(1-2/3)
= 2(1- (2/3)^(n-1) )
an = 3 - 2(2/3)^(n-1)
(2)
nan = 3n - 2[n(2/3)^(n-1)]
let
S = 1.(2/3)^0+2.(2/3)^1+...+n(2/3)^(n-1) (1)
(2/3)S = 1.(2/3)^1+2.(2/3)^2+...+n(2/3)^n (2)
(1)-(2)
(1/3)S = [1+2/3+...+(2/3)^(n-1)]- n(2/3)^n
= 3[1- (2/3)^n] - n(2/3)^n
S =9[1- (2/3)^n] - 3n(2/3)^n
= 9 - (3n-9)(2/3)^n
Sn = 3n(n+1)/2 - 2S
=n(n+1) - 18 + 2(3n-9)(2/3)^n
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※...
答：a2-a1=2/3 累加 an -a1=2/3 +(2/3)²+...+(2/3)^(n-1)an=a1+2/3 +(2/3)²+...+(2/3)^(n-1)=1+2/3 +(2/3)²+...+(2/3)^(n-1)=1×[1-(2/3)ⁿ]/(1-2/3)=3 -3×(2/3)ⁿ数列{an}的通项公式为an=3-3×(2/3)&...
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属
答：{a(n+1) -an} 是等比数列, q=2/3 a(n+1) -an = (2/3)^(n-1).(a2-a1)= (2/3)^n bn = a(n+1) -an = (2/3)^n a(n+1) - an = (2/3)^n an -a(n-1) = (2/3)^(n-1)an - a1 = (2/3)+(2/3)^2+...+(2/3)^(n-1)= (2/3)(1- (2/3...
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※...
答：可以看出B1=A2-A1=2/3 Bn+1=An+2-An+1=5/3An+1 - 2/3An 可以得An=Bn+1-5/3Bn 代入Bn=Bn+2 - 5/3Bn+1 - Bn+1 + 5/3Bn = Bn+2 - 8/3Bn+1 + Bn 话说此题打字也很烦～～的确悬赏分有点少
设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an,(n属于N※...
答：即b(n+1)=(2/3)b(n)b(n)是等比数列 b1=a2-a1=5/3-1=2/3 首项为2/3 公比为2/3 b(n)=(2/3)^n n>=1 n是整数 (2)a(n)-a(n-1)=b(n-1)=(2/3)^(n-1)a(n-1)-a(n-2)=b(n-2)=(2/3)^(n-2)...a(3)-a(2)=b(2)=(2/3)^2 a(2)-a(1)=b(...
设数列{an}满足a1=1,a(n+1)-an=1/2^n,(1)求数列{an}的通项公式(2)令...
答：a（n-1）-a（n-2）=1/2^（n-2） 以此规律一直写下去写到最后一个是a2-a1=1/2 然后把所有的加起来可得an-a1=1/2+1/2^2+……1/2^（n-1）（里面共有n-1项） 所以an=2-1/2^（n-1）第二题bn= 2n-1/2^（n-1）*n 过程就不写了，就是一个等差数列+等比数列与...
设数列{an}满足a1=1,a2=4,a3=9,an=an-1+an-2-an-...
答：解：由an=an-1+an-2-an-3，得 an+1=an+an-1-an-2，两式作和得：an+1=2an-1-an-3．即an+1+an-3=2an-1（n=4，5，…）．∴数列{an}的奇数项和偶数项均构成等差数列，∵a2=4，a4=12，∴偶数项公差为8．则a2014=a2+8（1007-1）=4+8×1006=8052．故答案为：8052．
设数列{an}满足:a1=1,an+1=3an属于N+。求{an}的通项公式及前n项和Sn...
答：解：a(n+1)=3an a(n+1)/an=3为定值 所以{an}是以a1=1为首项，q=3为公比的等比数列 于是 an=a1xq^(n-1)=1x3^(n-1）=3^(n-1）Sn=a1(1-q^n)/(1-q)=(3^n-1)/(3-1)=(3^n-1)/2
有以下五个命题:(1)设数列{an}满足a1=1,an+1=2an+1,则数列{an}的通项...
答：对于（1）：∵a1=1，an+1=2an+1，∴an+1+1=2（an+1），又a1+1=2，∴数列{an+1}是以2为首项，2为公比的等比数列，∴an+1=2?2n-1=2n，即an=2n-1；∴数列{an}的通项公式为an=2n-1，（1）正确；对于（2）：△ABC中，∵a2+b2-c2＞0，∴角C为锐角，但△ABC不一定是锐角三角...
(Ⅰ)设数列{an}满足a1=1,an+1=an2+5,n=1,2,3,…,证明对所有的n≥1,有...
答：证明：（i）由数学归纳法知，an＞0，∴an+1=an2+5=an2+4+1≥2an×2+1=4an+1． （ii） 对k≥2，有ak=ak-12+5=ak-12+4+1＞4ak-1+1＞4（4ak-2+1）+1＞…＞4k-1a1+4k-2+…+4+1=4k? 13∴11+3ak＜14k． 对所有的n≥1，有11+3an≤14n∴11+3a1+11+3a2+…+...
设数列an满足a1=1,a(n+1)-an=1/2^n.(1)求数列an的通项公式,(2)令bn=...
答：an -a1 = 1/2+..+1/2^(n-1)an = 1+ 1/2+..+1/2^(n-1)= 2(1-(1/2)^n)bn = nan = 2n- n(1/2)^(n-1)Tn =b1+b2+..+bn = n(n+1) - summation(i:1->n) i.(1/2)^(i-1)consider 1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)1+2x+..+nx^(...
