Download K.P.Pandeyend :: Characteristics of triangular numbers

OCTOGON MATHEMATICAL MAGAZINE
Vol. 17, No.1, April 2009, pp 282-284
ISSN 1222-5657, ISBN 978-973-88255-5-0,
www.hetfalu.ro/octogon
282
Characteristics of triangular numbers
K.P.Pandeyend30
INTRODUCTION
In this paper we have studied about the characteristics of the triangular
numbers which are of course positive integers obtained by taking the sum of
consecutive positive integers starting from unity.
DISCUSSION
Theorem 1. The sum of any two consecutive triangular numbers is
necessarily a perfect square.
Proof. We have
Tn + Tn+1 =
n (n + 1) (n + 1) (n + 2)
(n + 1) (n + n + 2)
+
=
=
2
2
2
= (n + 1) (n + 1) = (n + 1)2
which is a perfect square for any positive integer n, hence the result.
Theorem 2. Any triangular number can never be expressed as the sum of
two consecutive squares.
n
o
Proof. Supposing contrary, let for any positive integer n, n2 + (n + 1)2 is
a triangular number, then definitely
n
o
8 n2 + (n + 1)2 + 1
must be a perfect square,
n
o
⇒ 8 n2 + (n + 1)2 + 1 = m2 (say)
⇒ 8 n2 + n2 + 2n + 1 + 1 = m2
⇒ 16n2 + 16n + 9 = m2
⇒ (4n + 2)2 + 5 = m2
⇒ (4n + 2)2 = m2 − 5
=9−5=4
for m = 3,
⇒ 4n + 2 = 2 ⇒ n = 0
30
Received: 17.03.2009
2000 Mathematics Subject Classification. 05A16
Key words and phrases. Triangular numbers
Characteristics of triangular numbers
283
which is a contradiction of the fact that n > 0, hence the result.
Theorem 3. The product of any two consecutive triangular numbers can
never be a perfect square.
Proof. For any positive integer n, we have
(n + 1)2 n2 + 2n
n (n + 1) (n + 1) (n + 2)
Tn Tn+1 =
=
=
2
2
4
n
o
(n + 1)2 (n + 1)2 − 1
=
4
n
o
due to the term (n + 1)2 − 1 , the R.H.S. can never be a perfect square for
any positive integer n, hence the result.
Theorem 4. The product of any two consecutive triangular numbers is just
the half of another triangular number.
Proof. Let n be any positive integer, then we have
(n + 1)2 n2 + 2n
n (n + 1) (n + 1) (n + 2)
Tn Tn+1 =
=
=
2
2
4
(
2
)
n + 2n + 1 n2 + 2n
n2 + 2n + 1 n2 + 2n
1
=
=
=
4
2
2
=
1
1 m (m + 1)
, where n2 + 2n = m (say) Tm ,
2
2
2
hence the result.
Theorem 5. Product of any two alternate triangular numbers is just the
double of another triangular number.
Proof. Let n be any positive integer, then
n (n + 1) (n + 2) (n + 3)
n (n + 3) (n + 1) (n + 2)
=
=
2
2
4
2
n2 + 3n n2 + 3n + 2
n + 3n
n2 + 3n
=
+1 =
=
4
2
2
n 2
o
 2
n +3n
2

 n +3n
+
1
2
2
n + 3n
= m.
= 2Tm , where
=2


2
2
Tn Tn+2 =
Hence the result.
284
K.P.Pandeyend
Theorem 6. For n ≥ 2, we have
2
2
− Tn−1
T(n+1)2 − Tn2 = Tn+1
Proof. We have
T(n+1)2 − Tn2 =
n
o
(n + 1)2 (n + 1)2 + 1
2
n2 n2 + 1
−
=
2
2 2n3 + 3n2 + 3n + 1
n2 + 2n + 1 n2 + 2n + 2 − n2 n2 + 1
=
=
=
2
2
= 2n3 + 3n2 + 3n + 1
And,
2
Tn+1
−
2
Tn−1
=
n2 − n
−
4
Thus
(n + 1) (n + 2)
2
2
2
−
(n − 1) n
2
2
=
(n + 1)2 (n + 2)2
−
4
4 2n3 + 3n2 + 3n + 1
=
= 2n3 + 3n2 + 3n + 1
4
2
2
− Tn−1
.
T(n+1)2 − Tn2 = Tn+1
Hence the result.
Prof & Head of Applied Mathematics
Radharaman Institute of Technology and Science
Ratibad-Bhopal, India.
e-mail: [email protected]