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# X n=0 xn n X k=0 2n + 1 n − k  y2k+1

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Problem 12160

(American Mathematical Monthly, Vol.127, February 2020) Proposed by H. Ohtsuka (Japan) and R. Tauraso (Italy).

LetFn be thenth Fibonacci number, and let Ln be thenth Lucas number. Prove

n

X

k=0

2n + 1 n − k



F2k+1= 5n and

n

X

k=0

2n + 1 n − k



L2k+1=

n

X

k=0

2k k

 5n−k

for alln ∈ N.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Let

f (x, y) :=

X

n=0

xn

n

X

k=0

2n + 1 n − k



y2k+1=

X

k=0

y2k+1

X

n=k

2n + 1 n − k

 xn

= y

X

k=0

(xy2)k

X

n=0

2n + (2k + 1) n

 xn

= 2y

√1 − 4x(1 +√ 1 − 4x)

X

k=0

 4xy2 (1 +√

1 − 4x)2

k

= 2y

√1 − 4x(1 +√

1 − 4x)

1 − (1+4xy1−4x)2 2



= y(1 +√

1 − 4x)

√1 − 4x 1 +√

1 − 4x − 2x(1 + y2) where we used the known identity

X n=0

2n + m n



xn= 1

√1 − 4x

 2

1 +√ 1 − 4x

m

(see (5.72) in Concrete Mathematics by D. E. Knuth, O. Patashnik, and R. Graham).

Since

F2k+1=a2k+1− b2k+1

√5 and L2k+1= a2k+1+ b2k+1 with a := (1 +√

5)/2 and b := (1 −√

5)/2, it follows that X

n=0

xn

n

X

k=0

2n + 1 n − k



F2k+1= f (x, a) − f(x, b)

√5 ,

and

X

n=0

xn

n

X

k=0

2n + 1 n − k



L2k+1= f (x, a) + f (x, b).

On the other hand

g(x) :=

X n=0

5nxn = 1 1 − 5x. and

h(x) :=

X n=0

xn

n

X

k=0

2k k



5n−k= 1

√1 − 4x (1 − 5x). Hence, in order to show our identities, it remains to check that

f (x, a) − f(x, b) =√

5 g(x) and f (x, a) + f (x, b) = h(x)

which can be easily verified. 

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