# Video: Finding a Certain Term in a Binomial Expansion

Find the term independent of π₯ in the expansion of (π₯ + 1/π₯)ΒΉΒ² β (π₯ β 1/π₯)ΒΉΒ².

03:28

### Video Transcript

Find the term independent of π₯ in the expansion of π₯ plus one over π₯ all to the power of 12 minus π₯ minus one over π₯ all to the power of 12.

We have an expression that needs expanding. And weβre looking to find the term thatβs independent of π₯. Another way of describing this term is to think of it as the constant term. And this will exist roughly in the middle of each of the expansions. And letβs see why. The binomial expansion of π plus π to the power of π is given as π to the power of π plus π choose one multiplied by π to the power of π minus one multiplied by π plus π choose two multiplied by π to the power of π minus two multiplied by π squared, all the way through to π to the power of π.

So letβs see what this looks like when we have the expansion of π₯ plus one over π₯ to the power of 12. Our π here is π₯. Our π is one over π₯. And π is equal to 12. So the first term is quite simple in this expansion. Itβs just π₯ to the power of 12. Our second term is 12 choose one multiplied by π₯ to the power of 11 multiplied by one over π₯ to the power of one. Then we have 12 choose two multiplied by π₯ to the power of 10 multiplied by one over π₯ squared and so on. Now, since the value of π is even, this means thereβs going to exist a term for which the power of π₯ is equal to the power of one over π₯. In fact, theyβre both going to be six.

This is the seventh term of this sequence. Itβs given by 12 choose six multiplied by π₯ to the power of six multiplied by one over π₯ again to the power of six. One over π₯ all to the power of six is one to the power of six over π₯ to the power of six which is just one over π₯ to the power of six. And then letβs look at what happens. π₯ to the power of six multiplied by one over π₯ to the power of six is the same as π₯ to the power of six divided by π₯ to the power of six which is simply one. And so the term independent of π₯ in the expansion of π₯ plus one over π₯ all to the power of 12 is 12 choose six.

Weβre not going to evaluate 12 choose six just yet. Letβs repeat this process for the expansion of the second bracket. Our first few terms look extremely similar. But instead of one over π₯, we have negative one over π₯ in our brackets. We are, however, still interested in the seventh term. Itβs 12 choose six multiplied by π₯ to the power of six multiplied by negative one over π₯ to the power of six. But in fact, negative one over π₯ to the power of six is just one over π₯ to the power of six. Remember, when a negative number is raised to a positive exponent or a positive power, it becomes a positive number.

Once again, π₯ to the power of six multiplied by one over π₯ to the power of six is one. And we see that the seventh term in expansion of π₯ minus one over π₯ to the power of 12 and the constant of the term independent of π₯ is 12 choose six. Weβre subtracting the expansion of π₯ minus one over π₯ to the power 12 from the expansion of π₯ plus one over π₯ to the power of 12. So to find the term independent of π₯ in this expansion, we work out 12 choose six minus 12 choose six which is, of course, zero. The term independent of π₯ is zero.