To find such a method, we will first look for patterns which might help us. Some of the most obvious patterns we notice are related to the diagonals: the coefficients of the terms in the first diagonal only contain ones, whereas the coefficients in the second diagonal contain consecutive integers. Hence, their sum is equal to 4. Finally, we see there is a relation between the coefficients on consecutive rows: if we add the two coefficients in the row above, we get the coefficient in the following row.
Although, in much of the Western world, the triangle is named after the French mathematician Blaise Pascal, it was, in fact, well known to mathematicians centuries before in places such as China, Persia, and India.
Later, we will see how its properties give us a method to expand general binomials. Then, since all rows start with the number 1, we can write this down. We can then add each consecutive pair of elements of the sixth row and write their sum in the gap beneath them. We will demonstrate this process below. Starting with the first pair of terms, 1 and 5, we add them together to get 6 and place it into the space gap beneath them.
Therefore, starting from the first and second rows, which only contain ones, we can create the third rows by adding consecutive terms, as shown in the figure below. Similarly, we can write the other rows using the same method, until we get to the sixth row. When working with binomial expansions we might only be interested in specific terms, or even the coefficients of a specific terms. Individual terms can often be identified by looking for powers of a certain variable.
If however, both terms in a binomial contain the same variable, it might not be clear which powers of a certain variable will appear. Given that we have the product of two binomials raised to a power, it is usually helpful to expand each set of parentheses separately; then, we can consider their product.
Note that the single entry at the top of the triangle is conventionally referred to as the 0th row. The remainder of this explainer will seek to provide some tricks reduce the amount of calculation required for such questions. We might also choose to directly use the terms in the original binomial, and incorporate the coefficients found using Pascals triangle. At this point we could simplify our terms and reach an answer, however for binomials with large powers, notice that writing out such an expansion can be quite laborious!
Instead of writing out the entire expansion, this method gives us a shortcut to find individual terms. Fortunately, this would not have changed our answer. When we expand a binomial with a "—" sign, such as a — b 5 , the first term of the expansion is positive and the successive terms will alternate signs.
With all this help from Pascal and his good buddy the Binomial Theorem, we're ready to tackle a few problems. Take a look at Pascal's triangle. From the fourth row, we know our coefficients will be 1, 4, 6, 4, and 1. That negative sign means that the first term of our expansion will be positive, and the following terms will alternate signs. The exponents will start at x 4 y 0 and move to x 3 y 1 , etc. Wasn't that much easier than trying to multiply the expression out? We probably wouldn't have even tried, no matter how much the polynomials complained to us.
Have Pascal, will travel all the way to the answer. Our coefficients are 1, 5, 10, 10, 5, 1. The 2 n doesn't change anything; treat it like a single term.
With Pascal on our side, we feel invincible. However, that isn't quite the case. We're still weak to bullets, being dipped in lava while a villain monologues at us, or a well-timed chocolate. We also have trouble dealing with very large exponents; for every row we add to Pascal's triangle, it takes longer and longer to find the next. We also can't afford to skip any rows, or the usher will escort us out. Often times, when we have a binomial with a huge exponent, we only need one or a few terms.
For those problems, we'll use a special formula. It allows us to find the coefficient of just one specific term without finding any of the others.
Whew, that thing looks beastly. Just dreadful. It has fractions and exponents and factorials. We haven't even mentioned yet that the formula doesn't even tell you the sign of the term, either. If the binomial has a "—" sign, then the term is negative if r is even. If r is odd, then the term is positive. Maybe it won't be so bad once we solve a problem or two with it. Don't judge a book-sized formula by its cover, eh?
That binomial would take forever to multiply out, and we'd make hundreds of mistakes while doing it anyway. We're convinced already; give us the new formula, please. The 12th term is even, so our term will be negative. Remember, a factorial is a shortcut for writing a bunch of multiplication, like 3! That means we can cancel out a lot of that 14! As long as we know the formula, it works out.
This is it—the last problem before the summit of Polynomial Mountain. We've come a long way, so let's push just a little farther. The term will be positive because we've got a plus sign in our binomial, so all the terms are gonna be positive. Just think of how many terms we'd have to add together to get a number like 12, out of Pascal's Triangle. With that out of the way, we've made it to the top.
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