# Video: Finding a Certain Term in a Binomial Expansion

In the expansion of (6𝑎² + (1/𝑎))⁷, determine the term that has 𝑎².

04:09

### Video Transcript

In the expansion of six 𝑎 squared plus one over 𝑎 to the power of seven, determine the term that has 𝑎 squared.

So in order to solve this problem, actually find which term has 𝑎 squared, what I’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 𝑟 plus one, this is equal to 𝑛 choose 𝑟 multiplied by 𝑎 to the power of 𝑟 multiplied by 𝑏 to the power of 𝑛 minus 𝑟.

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 𝑛s and 𝑟s and our 𝑎s and our 𝑏s are. Well, our 𝑎 is our first term. So it’s gonna be six 𝑎 squared. Our 𝑏 is our second term, which is one over 𝑎 but remembering it’s positive. So it’s positive one over 𝑎. Our 𝑛 is seven because actually it’s the exponent of our parentheses. And so we’ve got that. And then, finally, 𝑟 is what we’re trying to find out because we want to determine the term that has 𝑎 squared.

Okay great. So now, we have the parts we need. Let’s actually substitute them into our formula. So this gives us that our term is equal to seven choose 𝑟 multiplied by six 𝑎 squared to the power of 𝑟 multiplied by one over 𝑎 to the power of seven minus 𝑟.

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

So the first rule we used was 𝑎 to the power of 𝑏 to the power of 𝑐 is equal to 𝑎 to the power of 𝑏𝑐. So we multiplied the exponents. And we actually used that to give us 𝑎 to the power of two 𝑟. And then, we used the rule that one over 𝑎 to the power of 𝑏 is equal to 𝑎 to the power of negative 𝑏 to give us 𝑎 to the power of negative seven minus 𝑟.

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

And now, because they all have the same base, which is 𝑎, we can actually equate the exponents. So this gives us two 𝑟 plus — and then we’ve got negative seven minus 𝑟 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 𝑟 minus seven plus 𝑟 equals two. So therefore, if we add seven to each side, we get three 𝑟 is equal to nine. Then divide by three, we get 𝑟 is equal to three.

So great, we’ve actually now found our 𝑟-value. So we can substitute it back into our formula to find out which term has 𝑎 squared. So then, if we substitute our 𝑟 back in, we’re gonna get 𝑇 three plus one. So it’s gonna be our fourth term. So this is gonna be equal to seven choose three multiplied by six 𝑎 squared to the power of three multiplied by one over 𝑎 to the power of seven minus three, which is gonna give us that the fourth term is equal to 35 multiplied by 216𝑎 to the power of six multiplied by 𝑎 to the power of negative four.

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