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ELEMENTARY PROBLEMS AND SOLUTIONS
Edited by A.P. HILLMAN
University of Santa Clara, Santa Clara, California
Send all communications regarding Elementary Problems and
Solutions to Professor A. P. Hillman, Mathematics Department, University of Santa Clara,
Santa Clara,
California.
Any problem be-
lieved to be new in the area of recurrent sequences and any new approaches to existing problems will be welcomed.
The proposer should
submit each problem with solution in legible form, preferably typed in
double spacing with name and address of the proposer as a heading.
Solutions to problems listed below should be submitted on separate signed sheets within two months of publication.
B-58
Proposed by Sidney Kravitz, Dover, New Jersey
Show that no Fibonacci number other than 1, 2, or 3 is equal to
^ Lucas number.
B-59
Proposed by Brother V. Alfred, St. Mary's College,
California
Show that the volume of a truncated right circular cone of slant
height
&
F
n
with
F
, and F . , the diameters of the bases is
n-1
n+1
^
B-60
F
n
+
l
-
F
n-1>/24 •
Proposed by Verner E. Hoggatt, Jr., San Jose State College, San Jose,
2
Show that
L0 L„ LO - 5F 0 , , = 1, where F
and
2n 2n+2
2n+l
n
n-th Fibonacci number and Lucas number,respectively.
B-61
L are the
n
Proposed by J. A. H. Hunter, Toronto, Ontario
Define a sequence
U,, U?, . . .
by U, = 3 and
U = U , + n 2 + n + 1 for
n
n-1
Prove that
B-62
California
n >1
.
U = 0 (mod n) if ng 0 (mod 3).
Proposed by Brother U. Alfred, St. Mary's College,
California
Prove that a Fibonacci number with odd subscript cannot be represented as the sum of squares of two Fibonacci numbers in more than
one way.
74
1965
E L E M E N T A R Y P R O B L E M S AND SOLUTIONS
B - 6 3 An old problem whose source is unknown, suggested
75
by Sidney Kravitz, Dover, New Jersey
In A ABC let s i d e s AB and AC be equal. Let t h e r e be
point D on side AB such that AD = CD = BC. Show that
2cos 3-A = A B / B C = (1 +
N/5)/2
a
.,
the golden m e a n .
SOLUTIONS
A BOUND ON BOUNDED FIBONACCI NUMBERS
B-44
Proposed by Douglas Lind, Falls Church,
Virginia
P r o v e that for e v e r y positive i n t e g e r k t h e r e a r e no m o r e than
k
k+1
n F i b o n a c c i n u m b e r s between n and n
Solution by the proposer.
A s s u m e the m a x i m u m ,
(1)
x
'
Now
nk < F
F ,9, . . . , F , < nk+1
r+2
r+n
xl,
r+1
n-1
**
r
r+n-1
r+j
**
=
F
P.- 2
r+2
r+n+l
But by (1),
n-1
2**
F
, • + F r+2
,7 >n-nk
r+j
and hence
r+n+1
thus proving the p r o p o s i t i o n .
»
k+1
F.
•
.
76
ELEMENTARY PROBLEMS AND SOLUTIONS
February
ANOTHER SUM
B-45
Proposed by Charles R. Wall, Texas Christian University,
Ft. Worth, Texas
Let H
be the n - t h teg e n e r a l i z e d Fibonacci n u m b e r , i. e. , let
n
H,1 and H~2 be a r b i t r a r y7 and H n+2
IO = H ., + H
n+1
n for n > 0. Show
+ Hn = Hn+4
that nHj + ( n - l ) H 2 + (n--2)H 3 +
J.-(n+2)H0-H1 ,
' 2 1
Solution by David Zeitlin,
Minneapolis,
Minnesota.
In B-20 (see Fibonacci Q u a r t e r l y ,
2(1964) p. 77), it was shown
that
n
1
H
j
= H
n+2 '
H
2
'
j=l
In B-40 (see Fibonacci Q u a r t e r l y , 2(1964), p . 155), Wall p r o p o s e d that
X
. . . + H, + H ,
JH. = (n + l)Hn+2_ - H n+4
1
2
Thus, the d e s i r e d sum
n
2
[(n+1) - j] H.. =
n
n
(n+1) 2 H. - 2
j=l
j=l
JH.
j=l
[(n+l)Hn+2 - (n+l)H2]- [(n+l)Hn+2-Hn+4+H1+H2]
=
H n + 4 - (n+2)H 2 - Hj
.
Also solved by Douglas Lind, Kenneth E. Newcomer, Farid K. Shuayto, Sheryl B.
Howard L. Walton, Charles Zeigenfus, and the proposer.
A CONTINUANT
B-46
DETERMINANT
Proposed by C.A. Church, Jr., Duke University,
Durham, North Carolina
Evaluate the n - t h o r d e r d e t e r m i n a n t
D
a+b
ab
0
0
1
a+b
ab
0
0
1
a+b
ab
= 1 0
0
1
a+b
Tadlock,