(Originally posted 2011-07-19.)

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.

### 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.

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