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I was solving my problem sheet and came upon this problem. original problem

I deduced the problem to this form, but I can't solve further. Can you please provide some help.

$$\sum_{r=0}^{12}\binom{13}{r}\binom{13}{r+1}$$

I tried to solve this sum by expanding the binomial coefficients inside the summation but I wasn't able to get to any conclusion.

I also thought to multiply two binomaials, but to no avail.

I was able to eliminate the options and get the answer from the original question. But I want to deduce an appropriate and proper solution to this summation.

Any insights or hint to help me solve this problem is much appreciated.

satyam singh
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Apr 02 '24 at 20:49
  • Is it better now @AnotherUser – satyam singh Apr 02 '24 at 20:55
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    A series is an infinite sum. That's simply a sum. – jjagmath Apr 02 '24 at 21:03
  • Take a look at https://oeis.org/A001791 – Jair Taylor Apr 02 '24 at 21:46

2 Answers2

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Your sum, $$\sum_{r=0}^{12}\binom{13}{r}\binom{13}{r+1}$$ evaluates to $9657700$.

We can get this answer using WolframAlpha or using Pascal's triangle. If you want to solve it by hand then you'll find out it's equal to $26 \choose 12$ or $26 \choose 14$.

You should clarify that the sum is not equal to $a+b$, it's equal to $ab$, so $ab = 9657700$. This isn't an answer, just a hint. Hope this helps :)

wasu
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  • Can you explain a bit further? I can't understand what are you trying to say. Also, Is there a short method to solve this sum without manual labour of finding each and every value. – satyam singh Apr 02 '24 at 21:11
  • https://math.stackexchange.com/questions/2208043/evaluate-the-sum-sum-k-1n-n-choose-k-1-n-choose-k – wasu Apr 03 '24 at 12:56
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It is known that: $$\binom{n}{k}$$ Corresponds to the $n$-th row and $k$-th column of the pascal triangle The entries of the $13$-th row are: $$F = \{1\;,\;13\;,\;78\;,\;286\;,\;715\;,\;1287\;,\;1716\;,\;1716\;,\;1287\;,\;715\;,\;286\;,\;78\;,\;13\;,\;1\}$$ So you are trying to find: $$\sum_{n=1}^{12}F_n\times F_{n+1} = 13+13\cdot 78+78\cdot286+\cdots+13\cdot 1= 2(13+13\cdot78+\cdots+1287\cdot1716)$$