FOURTH POWER FIBONACCI IDENTITIES FROM PASCAL'S TRIANGLE
VERNER E. HOGGATT, JR. and MARJORIE BICKNELL
San Jose State College, San Jose, California
In t h i s p a p e r ,
m a t r i x m e t h o d s a r e u s e d t o d e r i v e s o m e new-
fourth power Fibonacci identities.
We l e t
S be t h e 5X5 m a t r i x w h i c h
c o n t a i n s t h e f i r s t five r o w s of P a s c a l . ' s t r i a n g l e b e n e a t h a n d on i t s
secondary diagonal; that is,
0
0
0
0
0
0
0
0
1 3
0
1 2
1
1
'15
n
\i-lJ
P r o o f i s by i n d u c t i o n .
sn+1
1 4
6
1
S
4
1
= (s..)
i
n+1
Obviously,
1
3
1
T h e r i g h t c o l u m n e l e m e n t s of
0
-
a r e given by
1, 2, . . . ,
S has this form.
5.
Since
= (t..) = s s 1 1 ,
by d e f i n i t i o n of m a t r i x m u l t i p l i c a t i o n ,
rt
l5
t,c
25
tQC
35
-
F4
n+1 '
= 4F4
+ 4F F 3
n+1
n n+1
= 6F
4
n+1
4F
3
F
n+1
2
n+2 '
2
+ 12F F
+ 6F F
= 6F2
(F2
+ 2 F F ,, + F 2 )
n n+1
n n+1
n+1 n+1
n n+1
n
=6F2 ,F2
,
n+1 n+2
t.K
45
= 4F4,. +12F F3
+12F2F2,, + 4F3F
n+1
n n+1
n n+I
n n+1
= 4F
t -55
-
=
, , ( F 3 ,. + 3 F 2 , , F + 3 F , , F 2 + F 3 )
n+1 n+1
n+1 n
n+1 n
n
4
4
( F n + F n+1
, . ) * = F n+2
,9
.
261
4F
F
n+1
,
n+2
262
FOURTH POWER FIBONACCI IDENTITIES
December
Since only the recursion relation of the Fibonacci sequence was
used above, we have almost immediately a matrix identity for generalized Fibonacci numbers.
be the nth member of the generalto
n
ized Fibonacci sequence defined by
u, = a, u n = b, and u , , = u + u , .
J
^
1
2
n+1
n
n-1
Let U = (a..) be the column matrix defined by
11
Let u
5-i
u,
W '
i-1
U~>
2
.
,
, _
= 1, 2,
I
By our e a r l i e r proof, we can write
U
n+1
lu2
4u , , u i0
n+1 n+2
6u U
, 2 2
l 2
, Z
2
6u
u
n+1 n+2
3
4u u
l 2
4u
4u
snu = s n
4
l
U
n+1
-u 9
n+1 n+2
n+2
L
By the Cayley-Hamilton Theorem,
S must satisfy the matrix
equation
S n (S 5 - 5S 4 - 15S 3 + 15S2 + 5S - I) = 0 .
(1)
Consideration of Equation (1) leads us to the matrix equation
(!')
U
±c
n+5
- 5U ^A - 15U , . + 15U ± 7 + 5 U x l - U
=0
n+4
n+3
n+2
n+1
n
1
where
U is defined as the matrix S
U. Since the elements in the
n
first rows of the matrices of Equation (!') must also satisfy the r e cursion relation of (I 1 ), we have the identity
n+5
4
4
4
4
5(u
,A + 3u , Q - 3u , Q - u , , )
x
n+4
n+3
n+2
n+1'
Equation (1) can be rewritten as
(S - I) 5 = 25S2(S - I) .
It can easily be shown by induction that
1964
FROM PASCAL'S TRIANGLE
(S-I)4n+1
= 25nS2n(S-I)
263
.
We plan to investigate
(2)
(S - I ) 4 n + 1
= 2 5 n S 2 n (S - I) ,
(3)
(S - i ) 4 n + 2
= Z5 n S 2 n (S - I) 2 ,
(4)
(S - i ) 4 n + 3
=
(5)
(S - i ) 4 n + 4
= 2 5 n S 2 n (S - I ) 4 ,
2 5 n S 2 n (S - I ) 3 ,
F r o m Equation (2),
4n+l
j
S (S-I)
4n+1
=
+]v
/4
{ l)i
2
i=0
~
l
^
= 2 5 n S 2 n + J(S - I) .
) ^
i
I
Thus, equating e l e m e n t s in the upper right c o r n e r of t h e s e m a t r i c e s ,
4n+1
2
< '>
(
I
1
" >
/4 +l\
in
F
f+j = 2 5 n < F L j + l - F 2n +j > = A j •
i=0
S i m i l a r l y , from Equations (3), (4), and (5), we obtain
<3'> 2
^ ^ ^ U
i=0
\ i
= 25n(F
2n + j + 2- 2 F L + j+ l + F L + j )
/
= A j + 1 - A. ,
(4')
J^3
2
i-n
i=0
i/4n+3\
^ H
.
\ 1 /
F
4
i+j =
25
n
4
4
(F2n+j+3-3F2n+j+2
J+2
4n+4
(5')
Y
V>
2L
i=0
(-W4n+4W4
(
M
.
\
1
/
/
i+j
J+l
= *5n(F4
j+3
3F
4
4
,
2n+j+l-F2n+j)
J
-4F 4
* 2n+j+4
+
**2n+j+3
+6F4
*2n+j+2
4
4
- 4F
+F
)
j+2 r 2 nj ++lj + l j 2n+j ;
264
FOURTH POWER FIBONACCI IDENTITIES
December
But, also from (2),
2
/4n+5\
.
^H
-i-n
i=0
1
\
4
F
4
1 + j
2 5
=
4
< F 2n + j + 3 " F 2 n + j + 2 ) =
28,A.
/
so that we have the recursion relation
(6)
A. , A - 4A._L. + 6A. , 9 - 4A. , , + A. = 25 A. , 7 .
J+4
j+3
j+2
j+1
j
j+2
v
By use of well-known Fibonacci identities, we can rewrite Equations (2') and (4?) respectively to yield the following:
4n+l
2i=0
4 n + 3
Z
i=0
,. , -v
. /4n+l\
4
(_1)
I „ ) F i+j
X X
'
= 2 5
F
2n+j-l F 2n+j+2 F 2(2n+j)+l
'
. /4n + 3
(-1)M
F 4 , . = 2 5 n L7 , . L 7 , . , o F
i+j
2n+j 2n+j+3 2(2n+j)+3
\X x. / i+i
2n+i 2n+i+3
Equating elements in the first row of the column m a t r i c e s formed by
multiplying Equations (2) and (4) on the right by the m a t r i x
U,
and
1
taking
u, = 1, u ? = 3, we can rewrite Equations (2') and (4 ) to yield
the identities
4n+l
K7
Z
i=0
i/4n+I\
(- ) ( .
1
L
4
i+r
2 5
<5L2n+j-lL2n+j+2F2(2n+j)+l'
and
4n+3
. /4n+3\
v
' \
i=0
A
•
7
/
4
i+j
v
2n+j
2n+j+3
2(2n+j)+3'
for fourth powers of m e m b e r s of the Lucas sequence \L
I.
Returning to the recursion relation of Equation (6), we define
G(j)
= F
w/
By (6),
4
4
4
4
4
, . ^ - 4F 1 1 ± , - 1 9 F \ . ^ - 4F ± . x l + F _,_. .
n+j+4
n+j+3
n+j+2
n+j+1
n+j
1964
FROM PASCAL'S TRIANGLE
265
25 n (G(j+l) - G(j)) = A j + 4 - 4 A j + 3 - 1 9 A j + 2 - 4 A j + 1 + A. = 0 .
Thus,
G(j+1) - G(j) = 0, so that G(j) is a constant.
Taking n = j = 0,
G(j) = - 6 , leading to an identity given by Zeitlin in [ l ] .
If we redefine
G(j) by r e p l a c i n g the Fibonacci n u m b e r s by the c o r r e s p o n d i n g Lucas
n u m b e r s , we find that G(j) = -150. F u r t h e r , if we r e p l a c e m e m b e r s
of the F i b o n a c c i sequence by the c o r r e s p o n d i n g g e n e r a l i z e d F i b o n a c c i
2
we obtain G(j) = -6D , w h e r e D is the c h a r a c t e r i s t i c of
number,
the s e q u e n c e ,
D
=
U
2 " U l " U1U2 "
(See [2j and [3J for p r o p e r t i e s of the c h a r a c t e r i s t i c of F i b o n a c c i - t y p e
sequences.)
Finally, we d e r i v e a n o t h e r p r o p e r t y of the c h a r a c t e r i s t i c of a
s e q u e n c e . It is well-known that
^
?
?
F , ,T = F ,. + F . Define
2n+l
n+1
n
G(n)
F 0 .«, = 3F~ ., - F 0 ,
2n+3
2n+l
2n-l
and that
3F^n + F ^n-1.
n+1
Then,
G(n+1) + G(n)
so that
G(n)
-1)2.
= 0
F
- 3F
+ F
* 2n+3 - ^ 2 n + l
*2n-l
That i s ,
(-Dn2
3F2 + F 2 .
n
n-1
= F2
n+1
F o r g e n e r a l i z e d F i b o n a c c i n u m b e r s , it can be shown by induction that
n
R U
0
0
1
0
1
2
1
1
1
n
U
2u
2
l
lu2
2
u
L 2 J
n+1
2u , , u
n+1 n+2
n+2
The 3X3 m a t r i x R given above h a s been d i s c u s s e d in an e a r l i e r
a r t i c l e [ 4 ] . F r o m the c h a r a c t e r i s t i c equation of R, we obtain
266
FOURTH POWER FIBONACCI IDENTITIES
2
2
2
,
December
2
u . 0 - Zu I O - Zu , , + u
= 0n .
n+3
n+Z
n+1
n
Rewriting and defining
H(n),
H(n) .= u n+2
, 9 - 3un+1
, , + u n = -u n+3
. „ ~ un+1., + 3un+2. 0 = - H(n+1)
v
' .
Thus,
3u
H(n) = ( - l A u * where
l
+ u
0)
=
("1)n2(u2 - u i " uiu2^
D is the c h a r a c t e r i s t i c of the s e q u e n c e .
=
("1)n+1
2D
•
That i s ,
. , >nln
2
2 , 2
(-1) 2D
= u ,, - 3u + u ,
n+1
n
n~l
REFERENCES
1.
David
Zeitlin,
"On Identities for
Fibonacci
N u m b e r s , " The
A m e r i c a n M a t h e m a t i c a l Monthly, 70(1963), pp. 9 8 7 - 9 9 1 .
2.
B r o t h e r U. Alfred,
"On the O r d e r i n g of F i b o n a c c i Sequences, "
The F i b o n a c c i Q u a r t e r l y , 1:4, D e c , 1963, pp. 4 3 - 4 6 .
3.
A. P . Boblett, E l e m e n t a r y P r o b l e m B-29, The F i b o n a c c i
Quar-
t e r l y , 1:4, D e c , 1963, p. 75.
4.
V e r n e r E . Hoggatt and M a r j o r i e Bicknell, "Some New
Identities, " The F i b o n a c c i Q u a r t e r l y , 2(1964)
Fibonacci
F e b . , 1964, pp.
29-32.
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