Question

数列问题 求解 在线等

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.