could somebody please help me with this question, i really have no idea how to go about doing it? please?
thank you for your help
You are only asked to prove if it works for the first 4 natural numbers. So you wouldn't be completely hard done by to check each individual case!
$\displaystyle {2n \choose n} = \frac{(2n)!}{n!(2n-n)!} $
And
$\displaystyle \displaystyle \sum_{k=0}^{n} {n \choose k}^2 = {n \choose 0}^2 + {n \choose 1}^2 + {n \choose 3}^2+...+{n \choose n}^2 $$\displaystyle =\bigg(\frac{(n)!}{0!(n-0)!}\bigg)^2 + \bigg(\frac{(n)!}{1!(n-1)!}\bigg)^2+ \bigg(\frac{(n)!}{2!(n-2)!}\bigg)^2+...+\bigg(\frac{(n)!}{n!(n-n)!}\bigg)^2$
Just try both of these formula and see if they're equal. Once for n = 1, once for n =2, once for n = 3, and once for n = 4.