Question

数列求解问题

a(1)=1,a(2)=3,a(n)=a(n-1)+a(n-2)+1,求a(n)

Answer 1

a(n)=a+(n-1)d,

(1) a(1)=5=a,d为非负整数.
a(n)+a(n+1)+a(n+2)=3a+[n-1+n+n+1]d=3a+3nd=a+2a+3nd=a(m)=a+(m-1)d,
2a+3nd=(m-1)d,

2a+3nd=(m-1)d=10+3nd,
10=(m-1-3n)d,
d(>=0)是10的整数因子.
d的取值集合={1,2,5,10}.

(2)d不为0.
[a+(2m-1)d]^2 = [a(2m)]^2 = a(m+1)a(3m) = [a+md][a+(3m-1)d] = [a+(2m-1)d-(m-1)d][a+(2m-1)d+md] = [a+(2m-1)d]^2 + [a+(2m-1)d][-(m-1)d+md] - (m-1)md^2
=[a+(2m-1)d]^2 + d[a+(2m-1)d] - (m-1)md^2,
0=d[a+(2m-1)d] - m(m-1)d^2=d[a+(2m-1)d - m(m-1)d],
0=a+(2m-1)d - m(m-1)d=a+2md-d - m^2d + md = a+3md - (m^2+1)d,
0=2a+3(2m)d - [2(m^2+1) +1 - 1 ]d,
2a+3(2m)d = [2(m^2+1) + 1 - 1] d,
a(2m)+a(2m+1)+a(2m+2)=3a+[2m-1+2m+2m+1]d=3a+6md,
a[2(m^2+1)+1] = a + [2(m^2+1)+1-1]d = a+[2a+3(2m)d]=3a+6md=a(2m)+a(2m+1)+a(2m+2),
所以,
{a(n)}的第[2(m^2+1)+1]项可表示为a(2m)+a(2m+1)+a(2m+2),
{a(n)} 为"可拆数列".

(3)a(1)=2^k, k为正整数.
存在正整数m和正整数p,使得,
a(p)+a(p+1)+a(p+2)=3a+3pd=a(m)=a+(m-1)d,
2a+3pd=(m-1)d,
2^(k+1) = [m-1-3p]d,

d=2^(k+1)*[1/(m-1-3p)]
m>=2+3p时, d<=2^(k+1).
当m=2+3p时, d=2^(k+1)达到最大.

此时,a(n)=2^k + (n-1)2^(k+1),
s(n)=n*2^k + (n-1)n2^k=n^2*2^k,
s(k)=k^2*2^k,
a(k)=2^k+(k-1)2^(k+1)=[1+2(k-1)]2^k

0<200s(k)-[a(k)]^2 = 200k^2*2^k - [1+2(k-1)]^2*2^(2k),
0<200k^2 - [2k-1]^2*2^k,
0<200-[2-1/k]^2*2^k,
200>[2-1/k]^2*2^k,
25>[2-1/k]^2*2^(k-3),
k>3时,[2-1/k]^2*2^(k-3)单调递增.
k=6时,[2-1/6]^2*2^(6-3)=(11/6)^2*8=8*121/36=242/9=(250-8)/9>(250-25)/9=25,
所以,k>=6时,[2-1/k]^2*2^(k-3)>25.
k=5时,[2-1/5]^2*2^(5-3)=(9/5)^2*4=4*81/25=324/25<625/25=25.满足要求.
所以,1<=k<=5时,[2-1/k]^2*2^(k-3)<25,都满足要求.
因此,满足要求的最大的k=5.

Answer 2

an=a<n-1>+a<n-2>+1,①
∴a<n+1>=an+a<n-1>+1,②
②-①，a<n+1>-an=an-a<n-2>,
∴a<n+1>-2an+a<n-2>=0,
特征方程是x^3-2x^2+1=0,解得x1=1,x2=(1+√5)/2,x3=(1-√5)/2.
a1=1,a2=3,a3=5,
设an=x+y[(1+√5)/2]^n+z[(1-√5)/2]^n,则
1=x+[(1+√5)/2]y+[(1-√5)/2]z,
3=x+[(3+√5)/2]y+[(3-√5)/2]z,
5=x+(2+√5)y+(2-√5)z.
解得x=-1,y=(5+√5)/5,z=(5-√5)/5,
∴an=-1+[(5+√5)/5][(1+√5)/2]^n+[(5-√5)/5][(1-√5)/2]^n.