### Video Transcript

In the expansion of one plus π₯ to the power of 17, if the coefficient of the π plus the fourth term is equal to that of the two π plus threeth term, what is the value of π such that π is greater than one?

Before we can consider its coefficients, letβs begin by looking at the binomial expansion of one plus π₯ to the power of 17. The expansion of π plus π to the power of π is π choose zero multiplied by π to the power of π multiplied by π to the power of zero plus π choose one multiplied by π to the power of π minus one multiplied by π to the power of one, and so on. In our example, π is equal to one, π is equal to π₯, and π is equal to 17. And this general formula becomes 17 choose zero multiplied by one to the power of 17 multiplied by π₯ to the power of zero, and so on.

We can see though that the value of one to the power of any number is one. And this means that the coefficient of our first time is 17 choose zero. The coefficient of our second term is 17 choose one. And the coefficient of our third term is 17 choose two. We can generalize this and say that, in general, the coefficient of the πth term will be 17 choose π minus one. Letβs use this to find an expression for the coefficient of the π plus fourth term. Itβs 17 choose π plus four minus one, which is 17 choose π plus three. Weβll repeat this for the coefficient of the two π plus threeth term. Itβs 17 choose two π plus three minus one, which is 17 choose two π plus two.

This means weβre looking to find the value of π such that 17 choose π plus three is equal to 17 choose two π plus two. We are given that π is greater than one. So we could try substituting different values of π greater than one into our 17 choose π plus three and 17 choose two π plus two formulae.

However, there is a fact that we can use. We know that π choose π₯ is equal to π choose π¦ only when either π₯ is equal to π¦ or when π₯ plus π¦ is equal to π. Now, we know that the two terms that weβre looking for are different terms. So we canβt say that π plus three is equal to two π plus two. So as per the second part of our rule, weβll say that these two values, π plus three and two π plus two, add to make 17.

By collecting like terms, we see that three π plus five is equal to 17. And weβll solve this equation as normal. Weβll subtract five from both sides. And we get three π is equal to 12. Weβll then divide through by three. And that gives us an π-value as being equal to four. Letβs check this. 17 choose four plus three is equal to 17 choose seven. And 17 choose two π plus three, when π is equal to four, is 17 choose 10. 10 plus seven is indeed 17, as per our rule. And we can see then that π must be equal to four.