Video: Finding a Certain Term in a Binomial Expansion

Find π‘Žβ‚„ in the expansion of (5/√(π‘₯) + √(π‘₯)/5)⁹.

04:34

Video Transcript

Find the fourth term in the expansion of five over root π‘₯ plus root π‘₯ over five to the power of nine.

So, what we can use here is the binomial expansion and we have a general form for that. If we have π‘Ž plus 𝑏 to the power of 𝑛, then we can say that if we’re gonna expand this, it’s 𝑛 choose zero π‘Ž to the power of 𝑛 𝑏 to the power of zero plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 to the power of one et cetera up to 𝑛 choose 𝑛 π‘Ž to the power of zero 𝑏 to the power of 𝑛. So, what’s happening is the first term, you can see that, actually, the exponent is decreasing by one each time. And for the second term, the exponent is increasing by one each time.

So, in our question, what we’re looking for is π‘Ž sub four which is the fourth term. Well, we could do this a couple of ways. If we take a look at our general form for the binomial expansion, we could see that for the first term, it’s 𝑛 choose zero π‘Ž to the power of 𝑛 𝑏 to the power of zero. Then, the second term is 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 to the power of one. Well, if we take a look at our 𝑛 choose zero and 𝑛 choose one, then our 𝑛 is going to be nine because that’s the power that the parentheses are raised to. So, we’ve got nine. Then, we’ve got choose three. And that’s because the bottom number is always one less than the term number. And then, we’re gonna have π‘Ž to the power of six. And that’s because if we take a look at the expansion, the first π‘Ž would be π‘Ž to the power of nine. Then, we’d have π‘Ž to the power of eight, and then two further down the line would be π‘Ž to the power of six. And then, we’re gonna have 𝑏 to the power of three. And that’s because the exponent or power of 𝑏 is always one less than the term value. And also, as a quick check, if you add the exponents, they should add up to 𝑛. So, six add three is nine, yes, and 𝑛 is nine, great.

So now, let’s apply this to our expansion to find out what the fourth term is going to be. So, if we identify our π‘Ž and our 𝑏, our π‘Ž is five over root π‘₯ and our 𝑏 is root π‘₯ over five. So, what we’re gonna have is π‘Ž sub four or the fourth term is equal to nine choose three multiplied by five over root π‘₯ all to the power of six multiplied by root π‘₯ over five all to the power of three.

So, first of all, what we want to work out is what is the value of nine choose three. Well, nine choose three is equal to 84. And to find out what we can do is in our calculator, we press nine. And then, there’s an nCr button that you’ll find. Often, it’s you have to press shift and then a button to find nCr and then three. And that’ll give us 84. And then, what I’ve done for the other two terms is that I’ve changed them to exponent form. So, we’ve got five π‘₯ to the power of negative a half all to the power of six multiplied by π‘₯ to the power of a half over five all to the power of three. And we got that using a couple of exponent rules. And that is that if we have π‘₯ to the power of a half, it’s equal to root π‘₯. And if you have π‘₯ to the power of negative one, it’s equal to one over π‘₯.

So then, what we have is the fourth term is equal to 84 multiplied by 15625π‘₯ to the power of negative three. And that’s because we’ve raised five to the power of six, which is 15625. And then, we’ve got π‘₯ to the power of negative a half. Well, if we raise this to the power of six, then what we do is multiplying the exponents and we get negative three. And then, this is multiplied by π‘₯ to the power of three over two over 125. And as we’ve already said, it’s because we used another one of our exponent rules. And that is, if you’ve got π‘₯ to the power of π‘Ž to the power of 𝑏, then all we do is multiplying the exponents. So, we get π‘₯ to the power of π‘Žπ‘.

Okay, great. So now, what we need to do is another step of simplifying. So first of all, we’re gonna get 10500. And that’s because it’s 84 multiplied by 15625 divided by 125. And then, we’ll have π‘₯ to the power of negative three over two or negative three-halves. And that’s because we use one more exponent rule. And that is if we have π‘₯ to the power of π‘Ž multiplied by π‘₯ to the power of 𝑏 is equal to π‘₯ to the power of π‘Ž plus 𝑏. And if we have negative three and add on three over two or one and a half, we’re gonna get negative three over two.

So therefore, we can say that the fourth term of our expansion is going to be 10500π‘₯ to the power of negative three over two.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.