Combinatorics and Probability 751and this isF 2 n. The identity is proved.
(E. Cesàro)
862.Note that fork= 0 , 1 ,...,n,
(ak+ 1 +an−k+ 1 )(n+ 1 )= 2 Sn+ 1.If we add the two equal sums
∑
k(n
k)
ak+ 1 and∑
k(n
n−k)
an−k+ 1 , we obtain∑nk= 0(
n
k)
(ak+ 1 +an−k+ 1 )=
2 Sn+ 1
n+ 1∑nk= 0(
n
k)
=
2 n+^1
n+ 1Sn+ 1.The identity follows.
863.Newton’s binomial expansion can be used to express our sum in closed form as
Sn=1
4
[( 2 +
√
3 )^2 n+^1 +( 2 −√
3 )^2 n+^1 ].The fact thatSn=(k− 1 )^2 +k^2 for some positive integerkis equivalent to
2 k^2 − 2 k+ 1 −Sn= 0.View this as a quadratic equation ink. Its discriminant is
= 4 ( 2 Sn− 1 )= 2 [( 2 +√
3 )^2 n+^1 +( 2 −√
3 )^2 n+^1 − 2 ].Is this a perfect square? The numbers( 2 +
√
3 )and( 2 −√
3 )are one the reciprocal of
the other, and if they were squares, we would have a perfect square. In fact,( 4 ± 2
√
3 )
are the squares of( 1 ±
√
3 ). We find that=
(
( 1 +
√
3 )^2 n+^1 +( 1 −√
3 )^2 n+^1
2 n) 2
.
Solving the quadratic equation, we find that
k=1
2
+
( 1 +
√
3 )^2 n+^1 +( 1 −√
3 )^2 n+^1
22 n+^2
=1
2
+
1
4
[( 1 +
√
3 )( 2 +
√
3 )n+( 1 −√
3 )( 2 −
√
3 )n].This is clearly a rational number, but is it an integer? The numbers 2+
√
3 and 2−√
3
are the roots of the equation
λ^2 − 4 λ+ 1 = 0 ,