### 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 plus π to the power of π, then we can say that if weβre gonna expand this, itβs π choose zero π to the power of π π to the power of zero plus π choose one π to the power of π minus one π to the power of one et cetera up to π choose π π to the power of zero π to the power of π. So, whatβs 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 weβre looking for is π 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, itβs π choose zero π to the power of π π to the power of zero. Then, the second term is π choose one π to the power of π minus one π to the power of one. Well, if we take a look at our π choose zero and π choose one, then our π is going to be nine because thatβs the power that the parentheses are raised to. So, weβve got nine. Then, weβve got choose three. And thatβs because the bottom number is always one less than the term number. And then, weβre gonna have π to the power of six. And thatβs because if we take a look at the expansion, the first π would be π to the power of nine. Then, weβd have π to the power of eight, and then two further down the line would be π to the power of six. And then, weβre gonna have π to the power of three. And thatβs because the exponent or power of π is always one less than the term value. And also, as a quick check, if you add the exponents, they should add up to π. So, six add three is nine, yes, and π is nine, great.

So now, letβs apply this to our expansion to find out what the fourth term is going to be. So, if we identify our π and our π, our π is five over root π₯ and our π is root π₯ over five. So, what weβre gonna have is π 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, thereβs an nCr button that youβll find. Often, itβs you have to press shift and then a button to find nCr and then three. And thatβll give us 84. And then, what Iβve done for the other two terms is that Iβve changed them to exponent form. So, weβve 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, itβs equal to root π₯. And if you have π₯ to the power of negative one, itβs 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 thatβs because weβve raised five to the power of six, which is 15625. And then, weβve 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 weβve already said, itβs because we used another one of our exponent rules. And that is, if youβve got π₯ to the power of π to the power of π, then all we do is multiplying the exponents. So, we get π₯ to the power of ππ.

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