Video: Finding a Certain Term in a Binomial Expansion

In the expansion of (6π‘ŽΒ² + (1/π‘Ž))⁷, determine the term that has π‘ŽΒ².

04:09

Video Transcript

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

So in order to solve this problem, actually find which term has π‘Ž 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 π‘Ÿ plus one, this is equal to 𝑛 choose π‘Ÿ multiplied by π‘Ž to the power of π‘Ÿ multiplied by 𝑏 to the power of 𝑛 minus π‘Ÿ.

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 π‘Ÿs and our π‘Žs and our 𝑏s are. Well, our π‘Ž is our first term. So it’s gonna be six π‘Ž squared. Our 𝑏 is our second term, which is one over π‘Ž but remembering it’s positive. So it’s positive one over π‘Ž. Our 𝑛 is seven because actually it’s the exponent of our parentheses. And so we’ve got that. And then, finally, π‘Ÿ is what we’re trying to find out because we want to determine the term that has π‘Ž 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 π‘Ÿ multiplied by six π‘Ž squared to the power of π‘Ÿ multiplied by one over π‘Ž to the power of seven minus π‘Ÿ.

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 π‘Ž. So actually, we’ve written it as seven choose π‘Ÿ multiplied by six to the power of π‘Ÿ multiplied by π‘Ž to the power of two π‘Ÿ multiplied by π‘Ž to the power of negative seven minus π‘Ÿ. And in order to do this actually, we used a couple of exponent rules.

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

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

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

So great, we’ve actually now found our π‘Ÿ-value. So we can substitute it back into our formula to find out which term has π‘Ž squared. So then, if we substitute our π‘Ÿ 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 π‘Ž squared to the power of three multiplied by one over π‘Ž to the power of seven minus three, which is gonna give us that the fourth term is equal to 35 multiplied by 216π‘Ž to the power of six multiplied by π‘Ž to the power of negative four.

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

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