text: • Non ci sono risultati.
  
  # Show that for all positive integers m and n, n X k=1 (−4)−kn − k k − 1 3m X j=1 (−2)−jn + 1 − 2k j − 1
  
  N/A
  N/A
  Protected
  
  Condividi "Show that for all positive integers m and n, n X k=1 (−4)−kn − k k − 1 3m X j=1 (−2)−jn + 1 − 2k j − 1"
  
  Copied!
  2
  0
  0
  
  Testo completo
  
  (1)
  
  Problem 11914
  
  (American Mathematical Monthly, Vol.123, May 2016) Proposed by R. Chapman (UK) and R. Tauraso (Italy).
  
  Show that for all positive integers m and n,
  
  n
  
  X
  
  k=1
  
  (−4)−kn − k k − 1
  
  3m X
  
  j=1
  
  (−2)−jn + 1 − 2k j − 1
  
   m − k 3m − j
  
  
  
  = 0.
  
  (Here, xk = Qk−1i=0(x − i)/k! for x ∈ R).
  
  Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
  
  Let S(m, n) be the left-hand side then S(m, n) = −1
  
  2
  
  n
  
  X
  
  k=1
  
  (−4)−kn − k k − 1
  
  
  
  [x3m−1] (1 − x/2)n+1−2k(1 + x)m−k
  
  = −1 2
  
  X
  
  r+s=n−1,s≤r
  
  (−4)−1−sr s
  
  
  
  [x3m−1] (1 − x/2)r−s(1 + x)m−s−1
  
  =1 8
  
  X
  
  r+s=n−1,s≤r
  
  (−1)s2−r−sr s
  
  
  
  [x3m−1] (2 − x)r−s(1 + x)m−s−1 .
  
  For each positive integer m, introduce the generating function Fm(y) = 8
  
  X
  
  n=1
  
  S(m, n)(2y)n.
  
  Thus
  
  Fm(y) = 2y
  
  X
  
  r=0 r
  
  X
  
  s=0
  
  (−1)sr s
  
  
  
  [x3m−1] (2 − x)r−s(1 + x)m−s−1yr+s
  
  = 2y[x3m−1](1 + x)m−1
  
  X
  
  r=0
  
  (y(2 − x))r
  
  r
  
  X
  
  s=0
  
  r s
  
    −y
  
  (1 + x)(2 − x)
  
  s
  
  = 2y[x3m−1](1 + x)m−1
  
  X
  
  r=0
  
  
  
  (2 − x)y − y2 (1 + x)
  
  r
  
  = 2y[x3m−1] (1 + x)m−1 1 − (2 − x)y + y2/(1 + x)
  
  = 2y[x3m−1] (1 + x)m
  
  (1 + x) − (2 − x)(1 + x)y + y2
  
  = 2y
  
  (1 − y)2[x3m−1] (1 + x)m 1 + ux + (u2− u)x2
  
  (2)
  
  where we define u = 1/(1 − y). Now 1
  
  1 + ux + (u2− u)x2 = 1
  
  (1 − αx)(1 − βx) =
  
  X
  
  k=0
  
  akxk
  
  where
  
  ak= αk+1− βk+1
  
  α − β with α = −u +√
  
  4u − 3u2
  
  2 , β = −u −√
  
  4u − 3u2
  
  2 .
  
  In order to prove that Fm(y) = 0, and therefore S(m, n) = 0, we need to show that for each positive integer m,
  
  0 =
  
  m
  
  X
  
  j=0
  
  m j
  
  
  
  a3m−1−(m−j)=
  
  m
  
  X
  
  j=0
  
  m j
  
   α2m+j− β2m+j
  
  α − β = (α2(α + 1))m− (β2(β + 1))m α − β
  
  which holds because it is straightforward to verify that α2(α + 1) = u(α + 1)(β + 1) = β2(β + 1).
  
  
  
  Riferimenti
  
  Documenti correlati