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Binomial Theorem  Problem 3
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding a specified term in a binomial expansion. For this particular problem, we are only concerned with the fourth term we’re looking at when we expand this out. And so all that does is save a lot of work in terms of finding the other 9, 10 terms that are going to be involved in this equation.
Okay, so we know we would start out with x² to the tenth and our x² terms are solely going to decrease in exponent as our 2y terms increase. So to make our life a little bit easier, what I’m going to do is actually rewrite this as an addition sign so we don’t have to deal with any sign issues we might arise. So just x² and then this turns into a plus, minus 2y to the tenth. I didn’t really change the problem but I just sort of isolated the negative sign to the y term instead of having it as being a subtraction. And from here what we need to do is remember our binomial theorem, okay?
So we’re looking for the fourth term, so we start with x to the tenth, x to the ninth, x to the eighth, x to the seventh, and so we know that we are going to be left with x to the seventh. As we decrease our x terms our y terms are going to increase and our exponents, the degree on our terms is always going to add up to the same thing as we have here, which is the tenth.
Actually I made a mistake, this will be x but we’re actually dealing with x² so let’s erase that up and we’re actually going to end up with x² to the seventh and then 2y to the third. So no we also have to deal with the coefficient that we’re going to have, the n choose r term we had out here. And what, the cool thing about this is that it doesn’t really matter which one we have to choose. We have to choose one of these exponents and so what we end up with is 10 choose 7, we could equally do 10 choose 3 because they’re really the same thing, okay? If you remember this, this is 10 factorial, over 3 factorial. 10 minus 3 factorial, where this is 10 factorial divided by 7 factorial, 10 minus, oops! 7 factorial. Either way we end up with 10 factorial over 7 factorial 3 factorial.
So as long as you’re choosing your exponent as where your top number is and the bottom number is one of these two exponents, we’re going to be perfectly fine. Okay, so we know that this is our general fourth term. What we could do is expand this, or sometimes your teacher will be perfectly content with this. Let’s go through the process of expanding it just to make sure we have that practice down.
So we are looking at 10 to 7 which we have right here as 10 factorial over 7 factorial, 3 factorial. I know that my 7, 6, 5, 4 , 3, 2, 1s are all going to cancel, so what I’m going to be left with in the top is 10 times 9 times 8 over 3 factorial which is 3 times 2 times 1. Cancelling some terms, that goes to 3, that goes to 5, what we end up with is 5 times 3 times 8. 5 times 8 is 40 times 3 is 120. So we’re down to 120 times x² to the seventh. Power to power we multiply, so that turns into x to the fourteenth and 2y to the third, distribute that negative sign through will become 8y to the third. Combine like terms and 8 and our 120 multiply to be 960x to the fourteenth, y to the third.
So what we have done is using our binomial theorem found the specified term in our binomial expansion.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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