# Series identity with binomial coefficients [duplicate]

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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• I don't understand your question. Why do you think they're NOT equal? – Dr. belisarius Mar 5 '15 at 16:01
• I know they are equal. I need to prove it. – Marco Mar 5 '15 at 16:45
• – Sjoerd C. de Vries Mar 5 '15 at 18:56

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

• 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... – Marco Mar 6 '15 at 11:02