I can't get much else beyond what I wrote. I am sure that this problem is not secondary school level, it would be sixth form at least. Perhaps a better mathematician on the forums here can provide you with a method. I also found the difference between each term of the series you posted to be if its of any use.
This isn't a formal proof, but if you do casework for n=1, n=2, n=3, etc. then you find that the summation equals the middle number of every 2n-1th row in Pascal's Triangle. Thus, it fits the formula "2n-1 choose n-1" (because n is half of 2n).
(Original post by Edis)
This has nothing to do with the questions, this is the fifth page of the generating function.?
I'm sorry about that, I had a quick glance at what was written without actually reading it. Ashamed of that post. I would be keen on seeing a solution to your question myself. Sorry again, don't know what I was thinking.