 A recurrence relation is a natural way to describe the terms of a sequence, but even a simple recurrence relationship like poses a problem. Suppose you wanted to find the one hundredth term. Since we usually begin with n equals zero, the one hundredth term is the n equals ninety-nine term. And to find that, we need to know the n equals ninety-nine minus one ninety-eight and n equals ninety-nine times two ninety-seven terms. But wait, to find the n equals ninety-seven term, you'd need the n equals ninety-six and n equals ninety-five terms. And the n equals ninety-five term would require the n equals ninety-four, n ninety-three term, and lather, rins, repeat. We need to know all of the preceding terms, and if we made any mistake computing any term, every term after it would be wrong. What we want is a closed-form expression for the terms of a sequence. This would allow us to compute any term directly. If our sequence has a generating function, we could use the McLaren series for f of x to find the terms of the sequence. So if we assume a power series expansion for f of x, then our coefficients can be found. But this doesn't help since finding the nth derivative requires finding the first n derivatives. So how can we avoid finding the first n derivatives? We already know the McLaren coefficients for some series, one over one minus u, and e to the u. So if we can express our function in terms of either of these, we can find the coefficients directly. We'll start with the geometric series formula and get the exponential one later on. To motivate this, let's find the one-hundredth term of the recurrent sequence, and while we could compute a two, a three, a four, and so on up to a ninety-nine, this would take a lot of effort. Instead, let's find the generating function. So remember when finding the generating function, it helps to align on the xn terms, so we assume f of x has a powers series expansion. We want an a n minus one coefficient of x to the n. In our power series, we have a n minus one x to the power n minus one, so we'll take x times f of x. We'll also want a two a n minus two coefficient of x to the n. Again, in our power series, we have a n minus two x to the power n minus two, so we'll take two x squared f of x. Since the recurrence relationship is, subtracting gives us, since we know a zero and a one, we can substitute those in, hence all for f of x, hence our generating function will be. To make use of this, remember a series is a sum, is an integral. If we wanted to find the integral of generating function, we'd probably use the partial fraction decomposition, which will be, so if we add these two expressions, we get our generating function. So we know the series expansion for a function of the form one over one minus u, and so we find, and similarly, and since we can treat our power series as an infinite degree polynomial, we can add it term-wise. So our generating function expressed as a power series will be, and this gives us a closed form expression for our nth coefficient. So the one hundredth term will be where our index is equal to 99, remember zero is our first term, and so our formula tells us that it will be