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Topic:
Derivation of the Sum of a Series
Replies:
4
Last Post:
Sep 25, 2001 12:43 PM




Re: Derivation of the Sum of a Series
Posted:
Sep 24, 2001 10:37 AM


I don't mean to encourage more questions like Eric's but he does raise an interesting issue.
Consider the problem:
Find a polynomial in n whose value at n is the sum from i=1 to i=n of i^k, k a positive integer.
There are a couple of nice points you can make.
By a geometric argument you can see that this polynomial is less than n^{k+1}, for all n, hence must be a polynomial of degree less than or equal to k+1. For example, if k=1, then look at an n by n square. The terms of the sum are then the areas of squares of height i and base 1. If k=2, then the terms of the sums are volumes with base dimensions 1 and i and height i, and so on. This geometric argument can also be used to illustrate the usual derivation when k=1.
So what do you know about this polynomial P? Well, you know that
P(n)P(n1)=n^k
This is very powerful; two polynomials are equal iff all of their coefficients are equal.
You also know that P(0)=0. This means that the constant term of P is zero.
So look at a general polynomial of degree k+1, in n, evaluate the same polynomial at n1, subtract the second from the first, and set the coefficients equal to the coefficients of n^k. You can then work out the coefficients starting with the coefficient of n^{k+1}. You can do this for k=2, and have your students do it for k=3.
If there are other points you like to make with your students about this topic, Id be interested in hearing about it.
Best, Terry Gaffney

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