# Video: Finding a Certain Term in a Binomial Expansion

In the expansion of (6πΒ² + (1/π))β·, determine the term that has πΒ².

04:09

### Video Transcript

In the expansion of six π squared plus one over π to the power of seven, determine the term that has π squared.

So in order to solve this problem, actually find which term has π squared, what Iβm first gonna do is actually use a general term formula. Well, our general term formula actually tells us that if we have a term, which is π plus one, this is equal to π choose π multiplied by π to the power of π multiplied by π to the power of π minus π.

So now, when attempting to use this in this kind of question, the first thing that I need to do is actually decide what are πs and πs and our πs and our πs are. Well, our π is our first term. So itβs gonna be six π squared. Our π is our second term, which is one over π but remembering itβs positive. So itβs positive one over π. Our π is seven because actually itβs the exponent of our parentheses. And so weβve got that. And then, finally, π is what weβre trying to find out because we want to determine the term that has π squared.

Okay great. So now, we have the parts we need. Letβs actually substitute them into our formula. So this gives us that our term is equal to seven choose π multiplied by six π squared to the power of π multiplied by one over π to the power of seven minus π.

Okay great. So we have this. But what Iβm gonna do now is Iβm actually gonna split up a bit to help us actually look at the powers of π. So actually, weβve written it as seven choose π multiplied by six to the power of π multiplied by π to the power of two π multiplied by π to the power of negative seven minus π. And in order to do this actually, we used a couple of exponent rules.

So the first rule we used was π to the power of π to the power of π is equal to π to the power of ππ. So we multiplied the exponents. And we actually used that to give us π to the power of two π. And then, we used the rule that one over π to the power of π is equal to π to the power of negative π to give us π to the power of negative seven minus π.

Now if we look back at the question, what weβre looking at is the term that has π squared. So weβre really interested in just that term. So therefore, we can just take a look at our π terms because weβre just interested in the term where we have π squared. So in that case, we can say that π to the power of two π multiplied by π to the power of negative seven minus π is equal to π squared.

And now, because they all have the same base, which is π, we can actually equate the exponents. So this gives us two π plus β and then weβve got negative seven minus π equals two. And we get this because, in our first side of our equation, we multiplied two terms. So therefore, we add our exponents. So therefore, we have two π minus seven plus π equals two. So therefore, if we add seven to each side, we get three π is equal to nine. Then divide by three, we get π is equal to three.

So great, weβve actually now found our π-value. So we can substitute it back into our formula to find out which term has π squared. So then, if we substitute our π back in, weβre gonna get π three plus one. So itβs gonna be our fourth term. So this is gonna be equal to seven choose three multiplied by six π squared to the power of three multiplied by one over π to the power of seven minus three, which is gonna give us that the fourth term is equal to 35 multiplied by 216π to the power of six multiplied by π to the power of negative four.

So therefore, we can say that, in the expansion of six π squared plus one over π to the power of seven, the term that has π squared is equal to 7560π squared.