1 − 2z − z2+ z3 where x1:= sin(4π/7) sin(π/7

Problem 12117

(American Mathematical Monthly, Vol.126, May 2019) Proposed by M. Bataille (France).

Letn be a nonnegative integer. Prove that sinⁿ⁺¹(4π/7)

sinⁿ⁺²(π/7) −

sinⁿ⁺¹(π/7)

sinⁿ⁺²(2π/7)+ (−1)ⁿsinⁿ⁺¹(2π/7) sinⁿ⁺²(4π/7) is equal to

2√

7X(i + j + k)!

i!j!k! (−1)ⁿ⁻ⁱ2ⁱ

where the sum is taken over all triples(i, j, k) of nonnegative integers satisfying i + 2j + 3k = n.

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

Solution. Let ω := exp(2πi/7), then it is easy to verify that

(1 − x¹z)(1 − x²z)(1 − x³z) = 1 − 2z − z²+ z³ where

x1:= sin(4π/7)

sin(π/7) = sin(4π/7)

sin(6π/7)= ω²− ω⁻² ω³− ω⁻³

= 1 + ω + ω⁻¹,

x2:= sin(π/7)

sin(2π/7)= sin(6π/7)

sin(2π/7)= ω³− ω

−3

ω − ω⁻¹ = 1 + ω²+ ω⁻², x3:= −sin(2π/7)

sin(4π/7) = ω − ω⁻¹ ω²− ω⁻²

= 1 + ω³+ ω⁻³.

Hence, by the partial fraction decomposition, 1

1 − 2z − z²+ z³ = A1

1 − x¹z+ A2

1 − x²z + A3

1 − x³z, where

A1:= 1

(1 −^xx²1)(1 − ^xx³1) = x²1

2√

7 sin(4π/7),

A2:= 1

(1 −^xx¹2)(1 − ^xx³2) = − x²2

2√

7 sin(π/7),

A3:= 1

(1 −^x_x¹3)(1 − ^x_x²3) = x²3

2√

7 sin(2π/7). Finally

X

i+2j+3k=n

(i + j + k)!

i!j!k! (−1)ⁿ⁻ⁱ2ⁱ= [zⁿ]

∞

X

m=0

(2z + z²− z

3)^m

= [zⁿ] 1 1 − 2z − z²+ z³

= [zⁿ]

 A1

1 − x¹z+ A2

1 − x²z + A3

1 − x³z



= A1xⁿ1+ A2xⁿ2+ A3xⁿ3

= 1 2√

7

 sinⁿ⁺¹(4π/7) sinⁿ⁺²(π/7) −

sinⁿ⁺¹(π/7)

sinⁿ⁺²(2π/7)+ (−1)ⁿsinⁿ⁺¹(2π/7) sinⁿ⁺²(4π/7)



and we are done.