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I was solving a JEE practice problem and found a series I didn't know how to solve. The series was as follows:

C(15,15) + C(16,14) + C(17,13) + C(18,12) + ... + C(29,1) + C(30,0)

I'm not sure how to solve it, I would appreciate some help or just a tip to solve it.

Purv Kabaria
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    Surely you mean $C(30,0)$ at the end. Rather than for $n=15$, try finding $\sum\limits_{k=0}^n \binom{2n-k}{k}$ for some smaller values of $n$. They wind up being $2,5,13,34,\dots$ Those numbers should be familiar to you and appear in another famous sequence. Try to prove your claim if you want to be thorough, else just jump ahead to plugging in $15$ to get your final answer. – JMoravitz Apr 05 '22 at 18:10
  • or even more familiar do the intermediate values too: $2,3,5,8,13,\ldots$ – Henry Apr 05 '22 at 18:11
  • As for a hint as to prove the final answer, interpret this as counting the number of arrangements of dominos and monominos ($2\times 1$ blocks and $1\times 1$ blocks) in a row to get a total length of $30$, having broken apart into cases based on the number of total pieces (and as such, total number of dominos) used. This problem has a well known solution. – JMoravitz Apr 05 '22 at 18:14

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