Video: Finding a Certain Term in a Binomial Expansion

Find the term independent of π‘₯ in the expansion of (π‘₯ + 1/π‘₯)ΒΉΒ² βˆ’ (π‘₯ βˆ’ 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 we’re looking to find the term that’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 let’s see why. The binomial expansion of π‘Ž plus 𝑏 to the power of 𝑛 is given as π‘Ž to the power of 𝑛 plus 𝑛 choose one multiplied by π‘Ž to the power of 𝑛 minus one multiplied by 𝑏 plus 𝑛 choose two multiplied by π‘Ž to the power of 𝑛 minus two multiplied by 𝑏 squared, all the way through to 𝑏 to the power of 𝑛.

So let’s see what this looks like when we have the expansion of π‘₯ plus one over π‘₯ to the power of 12. Our π‘Ž here is π‘₯. Our 𝑏 is one over π‘₯. And 𝑛 is equal to 12. So the first term is quite simple in this expansion. It’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 𝑛 is even, this means there’s going to exist a term for which the power of π‘₯ is equal to the power of one over π‘₯. In fact, they’re both going to be six.

This is the seventh term of this sequence. It’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 let’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.

We’re not going to evaluate 12 choose six just yet. Let’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. It’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. We’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.

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