Here's another neat piece of algebra: A technique for summing series.

You know what b - a + c - b + d - c is. Right?

Suppose I were to write the same sum as:

(b - a) +

(c - b) +

(d - c)

The answer is still d - a. Right?

Now, suppose I re-label with s_{0} = a, s_{1} = b, s_{2} = c and s_{3} = d. We end up with:

(s_{1} - s_{0}) +

(s_{2} - s_{1}) +

(s_{3} - s_{2})

This is actually pretty scalable terminology as you can write s_{r} for any arbitrary value of r. And that's one of the strengths of algebra: generalisation.

So let's do that up to n:

(s_{1} - s_{0}) +

(s_{2} - s_{1}) +

...

(s_{n} - s_{n-1})

which is, of course, s_{n} - s_{0}.

But what has that got to do with summing series?

If we can replace each (s_{r} - s_{r-1}) by a single term u_{r} you may see the relevance...

u_{1} + u_{2} + ... + u_{n} = s_{n} - s_{0}.

The series summation boils down to a "simple" subtraction. The trick is to find these s terms, given the u terms. Let's try it with an example.

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### Summing The Integers

This is the series 1, 2, 3, ... , n.

The r'th u term is just r. u_{r} = r. So we now have to find the s_{r} term. Remember s_{r} - s_{r-1} has to equal r.

Try s_{r} = r (r + 1).

Then s_{r-1} = (r - 1) r or r (r - 1).

So s_{r} - s_{r-1} = [(r + 1) - (r - 1)] r or 2r. Not quite what we wanted. But we know - dividing by the factor of 2 - we should've guessed s_{r} = ½ r (r + 1).

So s_{n} - s_{0} = ½ n (n + 1) - ½ 0 (0 + 1) = ½ n (n + 1) - 0 = ½ n (n + 1).

The sum of the first n integers being ½ n (n + 1) is a well-known result. Admittedly it could've been done another way. But it's simple enough to show the method.

###

### Another Example - Summing The Squares Of The Integers

This is the series 1, 4, 9, ... , n² .

In this case we need to do something that will appear slightly perverse:

Rewrite r² as r ( r + 1) - r.

If you can split each of the terms in a series into two terms you can sum these sub terms. I just did the split. We already know how to sum the "r" portion. It's ½ n (n + 1). So we need to sum the r (r+1) portion and subtract ½ n (n+1) from the result.

Try s_{r} = r (r + 1) (r + 2).

Again we need to find s_{r-1}.

It's:

(r - 1) r (r + 1)

or, rearranging,

r (r+1) (r-1)

So

s

_{r} - s

_{r-1} = [(r + 2) - (r - 1)] r (r + 1) or 3r (r + 1).

This is 3 times what we want so we should've guessed s_{r} = 1/3 r (r + 1) (r + 2).

So this portion of the sum is 1/3 n (n + 1) (n + 2) - 1/3 0 (0 + 1)(0 + 2) or 1/3 n (n + 1) (n + 2).

But we need to subtract ½ n (n + 1) from this:

1/3 n (n + 1) (n +2) - ½ n (n + 1) = 2/6 n (n + 1) (n +2) - 3/6 n (n + 1)

or

1/6 n (n + 1) [ 2 (n + 2) - 3] = 1/6 n (n + 1)( 2 n + 1) .

If you try it for a few values you'll see it's right. This **isn't** such a well known result as for the sum of the integers.

I'm conscious there's been some fiddliness here - which is where **I** normally fall down.

But I think the "sum a series by converting it to a single subtraction" trick is a neat one - which is why I share it with you.

Tags:
summation
series
algebra
mathematics
maths