I have a following sum:
$S_g=\sum_{k=0}^g k\binom{4g+2}{2k}$
I can transform it into a different sum
$S_g=(2g+1)\sum_{k=1}^g\binom{4g+1}{2k-1}$
What is the closed form or what is the method to deal with any of above sums?
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$\begingroup$ This is a question for MSE, not MO. $\endgroup$– user64494Commented Feb 28, 2019 at 14:37
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$\begingroup$ Most questions related to partial sums of binomials I have come across was in MO. After your comment I checked for differences between MO and MSE. The question here, is the problem from my research, so it fits (though the solution appeared simple). $\endgroup$– piogorCommented Feb 28, 2019 at 14:51
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$\begingroup$ @user64494 It is arguably not appropriate to answer a question that you feel does not belong, because by so doing you are reinforcing the idea that inappropriate questions will be answered anyway. $\endgroup$– Todd TrimbleCommented Feb 28, 2019 at 18:34
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$\begingroup$ @Todd Trimble: I'd like to demonstrate Mathematica abilities in this field, no more and no less. $\endgroup$– user64494Commented Feb 28, 2019 at 19:03
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2 Answers
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According to Maple, $$ S_g = \left( g + \frac12\right) \left(16^g - {4 g \choose 2g}\right) $$
answered Feb 28, 2019 at 13:27
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1$\begingroup$ Thanks to your answer, I managed to find the proof. Please feel free, to include it in your post. Using $\binom{4g+1}{2k-1}=\binom{4g}{2k-1}+\binom{4g}{2k-2}$: $\sum_{k=1}^g \binom{4g+1}{2k-1}=\sum_{k=1}^g \binom{4g}{2k-1}+\sum_{k=0}^{g-1} \binom{4g}{2k}=\sum_{k=0}^{2g-1} \binom{4g}{k}=\dfrac{1}{2}(2^{4g}-\binom{4g}{2g})$ $\endgroup$– piogorCommented Feb 28, 2019 at 13:55
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$\begingroup$ This is $ a(n) = 2^{2n+1} $- $ {2n+1}\choose {n+1}$. How is this sequence related to the sum under consideration? $\endgroup$ Commented Feb 28, 2019 at 17:57
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$\begingroup$ $S_g = (2 g+1) A000346(2 g-1)$. $\endgroup$ Commented Feb 28, 2019 at 19:34
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$\begingroup$ Thank you. How is A000346 useful to the question under consideration? TIA. $\endgroup$ Commented Feb 28, 2019 at 19:49
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The Mathematica command
Sum[k*Binomial[4 g + 2, 2 k], {k, 0, g}]//FullSimplify
performs $$16^g g-\frac{\Gamma (4 g+3) \, _3F_2\left(2,1-g,\frac{3}{2}-g;g+\frac{5}{2},g+3;1\right)}{\Gamma (2 g-1) \Gamma (2 g+5)}.$$
