Video: EG19M2-ALGANDGEO-Q12

EG19M2-ALGANDGEO-Q12

05:08

Video Transcript

The ratio between the fifth term in the expansion of π‘₯ plus one over π‘₯ all to the power of 15 in descending powers of π‘₯ and the fourth term in the expansion of π‘₯ minus one over π‘₯ squared all to the power of 14 in descending powers of π‘₯ is negative one to 15. Find the value of π‘₯.

We’ll begin by recalling the general term of the expansion of a binomial. π‘Ž plus 𝑏 to the power of 𝑛 is equal to π‘Ž to the power of 𝑛 plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 and so on. We can also say that the general term π‘‡π‘Ÿ plus one, for π‘Ÿ is greater than or equal to zero and less than or equal to 𝑛, is 𝑛 choose π‘Ÿ multiplied by π‘Ž to the power of 𝑛 minus π‘Ÿ multiplied by 𝑏 to the power of π‘Ÿ.

Let’s use this to find the fifth term in the expansion of π‘₯ plus one over π‘₯ to the power of 15. In this case, π‘Ž is equal to π‘₯, 𝑏 is equal to one over π‘₯, 𝑛 is equal to 15, and then, since we’re finding the fifth term, π‘Ÿ plus one is equal to five. We can say then that π‘Ÿ must be equal to four. And the fifth term is, therefore, 15 choose four multiplied by π‘₯ to the power of 15 minus four multiplied by one over π‘₯ to the power of four. π‘₯ to the power of 15 minus four is the same as π‘₯ to the power of 11. And one over π‘₯ to the power of four is one to the power of four which is one over π‘₯ to the power of four.

And we can simplify this a little bit by dividing through by π‘₯ to the power of four. And we see that the fifth term in our expansion is 15 choose four multiplied by π‘₯ to the power of seven. Then for our second expansion, π‘Ž is equal to π‘₯, 𝑏 is equal to negative one over π‘₯ squared, 𝑛 is equal to 14, and then, since we’re finding the fourth term, π‘Ÿ plus one is equal to four which means π‘Ÿ is equal to three. And so the fourth term in this expansion is 14 choose three multiplied by π‘₯ to the power of 14 minus three multiplied by negative one over π‘₯ squared to the power of three.

π‘₯ to the power of 14 minus three is actually π‘₯ to the power of 11. And remember, when we have a negative number raised to an odd power, we still have a negative. So negative one over π‘₯ squared to the power of three is negative one over π‘₯ to the power of six. And once again, we can simplify a little by dividing through by π‘₯ to the power of six. And we get negative 14 choose three π‘₯ to the power of five. We are then told that the ratio between these two terms is negative one to 15. Dividing the second by the first gives us 15 divided by negative one. And that must be equal to negative 14 choose three multiplied by π‘₯ to the power of five over 15 choose four multiplied by π‘₯ to the power of seven.

We’ve cleared little bit of space to make room for our next bit of working out. And we should quite quickly see that we can multiply both sides by negative one. And that leaves us 14 choose three π‘₯ to the power of five over 15 choose four π‘₯ to the power of seven is equal to 15. And whilst we don’t want to evaluate fully 14 choose three and 15 choose four, we recall that 𝑛 choose π‘Ÿ is 𝑛 factorial over π‘Ÿ factorial multiplied by 𝑛 minus π‘Ÿ factorial. So this means that 14 choose three is 14 factorial over three factorial multiplied by 11 factorial. And 15 choose four is 15 factorial over four factorial multiplied by 11 factorial.

And rather than evaluating, we’re going to simplify this expression by recalling dividing by a fraction is the same as multiplying by its reciprocal. So 14 choose three divided by 15 choose four can be worked out by multiplying 14 choose three by the reciprocal of 15 choose four as shown. We can very quickly divide through by 11 factorial in these two examples. And then we recall that 15 factorial is 15 multiplied by 14 multiplied by 13 and so on or 15 multiplied by 14 factorial. And we can then divide through by 14 factorial.

14 factorial divided by itself is one. And then 15 factorial divided by 14 factorial is simply 15. We repeat this process with three factorial and four factorial. Four factorial is four multiplied by three factorial. So when we divide four factorial by three factorial, we’re simply left with four. And we can see that 14 choose three divided by 15 choose four simplifies to four fifteenths. π‘₯ to the power of five divided by π‘₯ to the power of seven simplifies to one over π‘₯ squared.

So we are now left with four fifteenths multiplied by one over π‘₯ squared all being equal to 15. To solve, we’re going to multiply both sides of this equation by π‘₯ squared to get four fifteenths is equal to 15π‘₯ squared. We then divide through by 15 to give us π‘₯ squared is equal to four over 15 squared. And at this stage, it’s important so that we could have evaluated 15 squared to be 225. But our next step is to square root both sides. This was unnecessary. When we do square root both sides, we see that π‘₯ is equal to plus or minus two fifteenths.

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