Hello,
I'm having trouble with question 7 in the link below. Can anyone help me with this question please(I have no idea how to do it)? Thanks
\[ \text{So upon equating coefficients and also doing some rearranging and factorising we obtain}\\ \binom{n}{r}^2(-1)^r+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{r-1}+(-1)^{r+1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{r-2}+(-1)^{r+2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^1 + (-1)^{2r-1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^0 +(-1)^{2r} \right] = \binom{n}{r}(-1)^r\]
Note that this equation is really long - when writing it down you'll likely need multiple lines.
\[ \text{We now divide both sides by }(-1)^r\text{ to obtain}\\ \binom{n}{r}^2+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}\\ \text{and moving that first }\binom{n}{r}\text{ term to the RHS we have}\\ \boxed{\binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}-\binom{n}{r}^2} \]
\[ \text{However since }(-1)^{-1} = (-1)^1,\text{ for any integer }k\text{ we also have }\boxed{(-1)^{-k} = (-1)^k}.\\ \text{This simplifies the LHS down considerably and gives us}\\ 2\left[ \binom{n}{r-1} \binom{n}{r+1} (-1)^1 + \binom{n}{r-2} \binom{n}{r+2} (-1)^2 + \dots + \binom{n}{1} \binom{n}{2r-1} (-1)^{r+1} +\binom{n}{0}\binom{n}{2r} (-1)^r \right] = \binom{n}{r} \left[1 - \binom{n}{r} \right] \]
You should be able to take it from here. Note that you need to divide \(-1\) on the LHS and RHS just once more.