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I have this apparently simple equation:

Sum[
  (p/(1 - p))^s*(q/(1 - q))^s*Binomial[n, s]*(Binomial[m - 1, s]*
    (p*q*(m + n) + (2*m - 1)*(-p - q + 1))), 
  {s, 0, n}] == 
Sum[
  (p/(1 - p))^s*(q/(1 - q))^s*Binomial[n, s]*
    ((-(-p - q + 1))*Binomial[m - 2, s] + m*p*q*Binomial[m, s] + 
       m*(-p - q + 1)*(Binomial[m - 2, s] + Binomial[m, s])), 
  {s, 0, n}]

Mathematica's FullSimplify command immediately tells me that this equation is an identity, giving me True as the output, but I fail to see the analytical reason.

All parameters are weakly positive and real, although I do not need to assume anything for Mathematica to tell me that it is indeed an identity.

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4
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Mar 5 '15 at 15:36
  • $\begingroup$ I don't understand your question. Why do you think they're NOT equal? $\endgroup$ – Dr. belisarius Mar 5 '15 at 16:01
  • $\begingroup$ I know they are equal. I need to prove it. $\endgroup$ – Marco Mar 5 '15 at 16:45
  • $\begingroup$ Or get-a-step-by-step-evaluation-in-mathematica. $\endgroup$ – Sjoerd C. de Vries Mar 5 '15 at 18:56
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It might help you to look at each side separately.

Starting with the left-hand side

lhs = Sum[(p/(1 - p))^s*(q/(1 - q))^s*
   Binomial[n, 
    s]*(Binomial[m - 1, s]*(p*q*(m + n) + (2*m - 1)*(-p - q + 1))), {s, 0, n}]

-Hypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + 2*mHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + pHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] - 2*mpHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p) (-1 + q))] + qHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)*(-1 + q))] - 2*mqHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p) (-1 + q))] + mpqHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)* (-1 + q))] + npqHypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)* (-1 + q))]

lhs = lhs // Simplify

(-1 + p + m*(2 + p*(-2 + q) - 2*q) + q + npq)* Hypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))]

And for the right-hand side

rhs = Sum[(p/(1 - p))^s*(q/(1 - q))^s*
   Binomial[n, 
    s]*((-(-p - q + 1))*Binomial[m - 2, s] + m*p*q*Binomial[m, s] + 
     m*(-p - q + 1)*(Binomial[m - 2, s] + Binomial[m, s])), {s, 0, n}]

-Hypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + mHypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + pHypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] - mpHypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + qHypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] - mqHypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + mHypergeometric2F1[-m, -n, 1, (pq)/((-1 + p)(-1 + q))] - mpHypergeometric2F1[-m, -n, 1, (pq)/((-1 + p)(-1 + q))] - mqHypergeometric2F1[-m, -n, 1, (pq)/((-1 + p)(-1 + q))] + mpqHypergeometric2F1[-m, -n, 1, (pq)/((-1 + p)*(-1 + q))]

rhs = rhs // Simplify

(-(-1 + m))(-1 + p + q) Hypergeometric2F1[2 - m, -n, 1, (pq)/((-1 + p)(-1 + q))] + m*(-1 + p)(-1 + q) Hypergeometric2F1[-m, -n, 1, (pq)/((-1 + p)(-1 + q))]

rhs = rhs // FullSimplify

(-1 + p + m*(2 + p*(-2 + q) - 2*q) + q + npq)* Hypergeometric2F1[1 - m, -n, 1, (pq)/((-1 + p)(-1 + q))]

The simplified forms are the same

lhs === rhs

True

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  • $\begingroup$ Thanks a lot for your help. It is very simple to get the algebrical expression of the simplified lhs (since the binomial coefficients are all the same as the ones of the resulting hypergeometric), whereas the rhs is what really struggles me. And "Trace" does not help... $\endgroup$ – Marco Mar 6 '15 at 11:02

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