Sums and Approximations So to compute the value of the annuity we need only evaluate this sum. We could plug in values for m, n, and p, compute each term explicitly, and then add them all up However, this particular sum has an equivalent"closed form"that makes the job easier. In general, a closed form is a mathematical expression that can be evaluated with a fixed number of basic operations(addition, multiplication, exponentiation, etc. )In contrast, evaluating the sum above requires a number of operations proportional to n 1.2 A Geometric sum Our goal is to find a closed form equivalent to the summation V m+ 1+p) 1+p(1+p)2 ...+ (1+p)k-1 This is a geometric sum, which means that consecutive terms all have the same ratio. In particular, the second term is 1/(1+p) times the first, the third is 1/(1+ p)times the second, and so forth. And weve already encountered a theorem about geometric sums Theorem 1. For all n> 1 and all z+1 This theorem can be proved by induction, but that proof gives no hint how the for- mula might be found in the first place. Here is a more insightful derivation based on the perturbation method. First, we let S equal the value of the sum and then"perturb"it by multiplying by a + 2+ + The difference between the original sum and the perturbed sum is not so great, because here is massive cancellation on the right side s-2S=1 Now solving for S gives the expression in Theorem 1 You can derive a passable number of summation formulas by mimicking the approach used above. Well look at some other methods for evaluting sums shortly� Sums and Approximations 3 So to compute the value of the annuity, we need only evaluate this sum. We could plug in values for m, n, and p, compute each term explicitly, and then add them all up. However, this particular sum has an equivalent “closed form” that makes the job easier. In general, a closed form is a mathematical expression that can be evaluated with a fixed number of basic operations (addition, multiplication, exponentiation, etc.) In contrast, evaluating the sum above requires a number of operations proportional to n. 1.2 A Geometric Sum Our goal is to find a closed form equivalent to the summation: �n−1 m m m m V = = m + + + . . . + (1 + p)k 1 + p (1 + p)2 (1 + p)k−1 k=0 This is a geometric sum, which means that consecutive terms all have the same ratio. In particular, the second term is 1/(1 + p) times the first, the third is 1/(1 + p) times the second, and so forth. And we’ve already encountered a theorem about geometric sums: Theorem 1. For all n ≥ 1 and all z = 1: �n−1 n k 1 − z z = 1 − z k=0 This theorem can be proved by induction, but that proof gives no hint how the for￾mula might be found in the first place. Here is a more insightful derivation based on the perturbation method. First, we let S equal the value of the sum and then “perturb” it by multiplying by z. S = 1 + z + z2 + . . . + zn−1 zS = z + z2 + . . . + zn−1 + zn The difference between the original sum and the perturbed sum is not so great, because there is massive cancellation on the right side: S − zS = 1 − z n Now solving for S gives the expression in Theorem 1: n 1 − z S = 1 − z You can derive a passable number of summation formulas by mimicking the approach used above. We’ll look at some other methods for evaluting sums shortly