Hey there!
For 7a) I'm not so sure how they're getting the bottom values for r in the coefficients... also, don't understand the restriction on r especially as I thought it could only be equal to 4.
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So this is the same trick as in the previous question, but this we are considering some unknown power of \(x\), \(x^r\). On the LHS, the coefficient is simply:
Now consider the right hand side:
So how do we get a power of \(x^r\)? Take each term from the right bracket and pair it off: \(1\) pairs with \(\binom{n}{r}\). \(4\) pairs with \(\binom{n}{r-1}\), and so on. You can see the result form from there.
As for the restriction, clearly \(r\le n+4\), just by the nature of the binomial expansion, so that is a start. However, for this result to be valid, \(r\) needs to be high enough that the bottom term of each of those binomial coefficients isn't negative - That is, we can't have \(\binom{n}{-2}\), for example. So, \(r\ge4\).