Video: MATH-ALG+GEO-2018-S1-Q15

In the expansion of (π‘₯Β² βˆ’ (1/π‘₯Β²))ΒΉΒ² in descending powers of π‘₯, find a term that contains π‘₯⁴. Then find the ratio between the coefficient of this term and that of the middle term.

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Video Transcript

In the expansion of π‘₯ squared minus one over π‘₯ squared all to the power of 12 in descending powers of π‘₯, find a term that contains π‘₯ to the power of four. Then find the ratio between the coefficient of this term and that of the middle term.

This question involves terms of the binomial expansion of π‘₯ squared minus one over π‘₯ squared all to the power of 12. We know that, in the general binomial expansion of π‘Ž plus 𝑏 all to the power of 𝑛, 𝑇 subscript π‘Ÿ plus one, the π‘Ÿ plus oneth term of this expansion, is 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the power of π‘Ÿ.

In our expansion, π‘Ž is π‘₯ squared, 𝑏 is negative one over π‘₯ squared, and 𝑛 is 12. We can substitute these for π‘Ž, 𝑏, and 𝑛 in the formula for the π‘Ÿ plus oneth term we have. To find that, the π‘Ÿ plus oneth term of our expansion is 12 choose π‘Ÿ times π‘₯ squared to the power of 12 minus π‘Ÿ times negative one over π‘₯ squared to the power of π‘Ÿ.

Now we can simplify this expression. We’ll leave the 12 choose π‘Ÿ as it is for the moment. But we can use the law of indices to simplify π‘₯ squared to the power of 12 minus π‘Ÿ. We know that π‘Ž to the power of 𝑏 all to the power of 𝑐 is π‘Ž to the power of 𝑏 times 𝑐. And so π‘₯ to the power of two to the power of 12 minus π‘Ÿ is π‘₯ to the power of two times 12 minus π‘Ÿ.

We can use another law of indices to simplify one over π‘₯ squared. One over π‘₯ to the 𝑛 is π‘₯ to the negative 𝑛. And so this becomes negative π‘₯ to the negative two, which we’re raising to the power of π‘Ÿ. we can simplify further. We can use the distributive property on the index of π‘₯. This becomes π‘₯ to the power of two times 12, which is 24, minus two times π‘Ÿ, which is two π‘Ÿ.

It’s also helpful to think of negative π‘₯ to the power of negative two all to the power of π‘Ÿ as negative one times π‘₯ to the negative two all to the power of π‘Ÿ. And then as π‘Ž times 𝑏 all to the power of 𝑐 is equal to π‘Ž to the power of 𝑐 times 𝑏 to the power of 𝑐. This is negative one to the power of π‘Ÿ times π‘₯ to the power of negative two to the power of π‘Ÿ. We write this in.

Now we’re almost done. This negative one to the power of π‘Ÿ doesn’t involve π‘₯. And so it’s part of the coefficient and should move to the front. 12 choose π‘Ÿ and π‘₯ to the power of 24 minus two π‘Ÿ we’ll leave as they are for the moment. We can, however, simplify π‘₯ to the power of two to the power of π‘Ÿ. Again, this is a power of a power. And so we see that this is equal to π‘₯ to the power of negative two π‘Ÿ.

Now we can write the product of two powers with the same base as a single power. We just add the indices. The index becomes 24 minus two π‘Ÿ minus two π‘Ÿ, which is 24 minus four π‘Ÿ. This is the π‘Ÿ plus oneth term of our expansion in simplest form.

Let’s clear some room so we can do something with it. Our first task from the question is to find the term that contains π‘₯ to the power of four. Looking at our expression for the π‘Ÿ plus oneth term of our binomial expansion, if this is some multiple of π‘₯ to the power of four, then our index, 24 minus four π‘Ÿ, must be four. 24 minus four π‘Ÿ equals four. 24 minus four π‘Ÿ equals four. We subtract 24 from both sides, finding that negative four π‘Ÿ is negative 20. And dividing through by negative four, we get that π‘Ÿ is equal to five.

We substitute this value of π‘Ÿ then, to find that the term containing π‘₯ to the power of four is 𝑇 subscript five plus one, which is negative one to the power of five times 12 choose five times π‘₯ to the power of 24 minus four times five. Of course, five plus one is six. Negative one to the power of any odd number is negative one. 12 choose five is 792. And unsurprisingly, π‘₯ to the power of 24 minus four times five is π‘₯ to the power of 24 minus 20, which is π‘₯ to the power of four. Simplifying then, we see that the term in our expansion containing π‘₯ to the four is negative 792 times π‘₯ to the four.

Now we move on to our next task, finding the ratio between the coefficient of this term and that of the middle term. First, we need to find the middle term. The index 𝑛 in our expansion is 12. And so the middle term is the term of order 12 over two plus one, which is the seventh term. You can see this because there are 12 plus one equals 13 terms of the expansion, corresponding to values of π‘Ÿ between zero and 12, which are the values for which 12 choose π‘Ÿ makes sense.

We work inwards to find the middle term. It is indeed term seven. As this is term seven, the value of π‘Ÿ is one less than this. It’s six. And so we substitute six for π‘Ÿ in the expression we have for term π‘Ÿ plus one. We get negative one to the power of six times 12 choose six times π‘₯ to the power of 24 minus four times six. Now negative one to the power of any even number is just one. 12 choose six is 924. And π‘₯ to the power of 24 minus four times six is π‘₯ to the power of 24 minus 24, which is π‘₯ to the power of zero. And anything to the power of zero is just one. So the middle term is 924.

Now remember, we’re looking for the ratio between the coefficient of the previous term and that of this middle term. The coefficient of the π‘₯ to the power of four term is negative 792. And we’re looking to divide this by the coefficient of this middle term. And as the term is a constant term, this coefficient is just the term itself, 924.

It might have been easier to see this had we kept π‘₯ to the power of zero as π‘₯ to the power of zero rather than simplifying it to one. So the ratio that we’re looking for is negative 792 over 924. And clearly, this can be simplified as both numerator and denominator are even. It turns out that these two coefficients have a lot more factors in common as their ratio when simplified is that of negative six to seven.

Now it’s not a coincidence that we get such a simple answer. We had defined the ratio of coefficients of two consecutive terms of binomial expansion, 𝑇 six and 𝑇 seven. And you might know that the ratio of π‘‡π‘Ÿ to π‘‡π‘Ÿ plus one is equal to π‘Ÿ over 𝑛 minus π‘Ÿ plus one times the coefficient of the first term of the binomial π‘Ž over the coefficient of the second term of the binomial 𝑏. This gives another method of finding the ratio required.

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