 The binomial theorem gives us a way to work with combinations. It's useful to keep in mind that in the binomial expansion of a plus b to the n, the sum of the exponents on a and b is n, and this product is multiplied by n choose i. So if we can rewrite a combinatorial expression in this form, we can interpret it as the power of a binomial. So for example, let's suppose we want to find the sum, 10 choose 0 plus 10 choose 1, and so on, up to 10 choose 10. So let's see if we can massage these terms so they look like the terms of the binomial expansion. So the binomial expansion of a plus b to the n, n choose k, is going to be multiplied by a power of a times a power of b, where the sum of the exponents is n. So remember, you can get anything you want as long as you pay for it, but it's also worth keeping in mind some things are free. And one of those free things is multiplication by 1. So if we want 10 choose 0 to be a power of 1 factor times a power of another factor, we can make both of those factors 1. So 10 choose 0 is 10 choose 0 times 1 times 1. Now with the binomial expansion, both of these are raised to a power where 1 power is the number of things we're choosing and the sum is the total number of things we're choosing from. So 1 factor of 1 will be raised to the 0 power, and the other will be raised to the power 10. And because 1 to any power is still 1, we maintain our equality. Likewise, 10 choose 1, we can rewrite that as 10 choose 1 times 1 times 1, or 10 choose 1 times 1 to the 9th, 1 to the 1st. And in fact, we can rewrite all of our terms this way, but this is just the expansion of 1 plus 1 to the 10th, which is, if we generalize the proceeding, we find that 1 plus 1 to the nth will be, and since 1 raised to any power is 1, and we can omit factors of 1 we get, and writing this in our summation notation, and the most important consequence is that this gives us a way to simplify the sum of binomial coefficients. The least important consequence is the theorem that for all whole numbers n, the sum of the binomial coefficients is 2 to power n. Or let's consider a different summation. Now, this one is in a more general form, so it's useful to keep in mind concrete doesn't hurt. Let's start with an example. So let's pick an arbitrary value for n, and the only real restriction is we don't want it to be too small because sometimes weird things happen for very small numbers, and it has to be something we can work with. So let's try n equals 4. So if n equals 4, this sum looks like, which looks like half of the binomial expansion, and in particular, it's missing the odd terms. In other words, we note that if n equals 4, the expansion of 1 plus 1 to the 8th will be, which includes all of the terms we want, plus some terms we don't want. To get rid of the odd terms, we'll invoke a very useful algebraic identity if we add a sum and a difference, we get twice one of the terms. So using the binomial theorem to expand 1 minus 1 to the 8th, well, that's really 1 plus negative 1 to the 8th, we note that the i-th term will be, so this will be positive for even i and negative for odd i. And so we find, and if we add them together we get, and we can divide everything by 2 to get... Now, we showed this is true for n equals 4, but we do want to generalize it, so remember, equals means replaceable. We start with n equals 4, and this means that everywhere we see a 4, we can replace it with n. And so we find all of our 4s... Well, we don't see any 4s, but remember that 8 came from 2 times 4, and our terms are 8 choose k, where k went from 0 to 8, not 8, 2 times 4. So we can replace 8 with 2n and get our general result, and we should try to answer this in the same format we started with, since the problem used our sigma notation, let's go ahead and rewrite the sum on the right in sigma notation, and we get... and we can always do a little calculus. So in something like this, we note that a coefficient like 10 choose 3 would appear in the expansion of... Oh, I don't know how about pi plus x to the 10th, so we'd get a bunch of terms. We get 10 choose 3, pi to the 7th, x to the 3rd, and if we differentiated, we'd get... Of course, we don't have to use pi as our constant. We could use any other non-zero constant. We'll have about 1 plus x to the 10th. We expand, differentiate both sides, and if we let x equals 1, we get...