Arithmetic This!

There’s a special name for this type of sequence:


Notice that each term differs from its preceding and succeeding term by a constant difference of 3 – this type of series can be represented as such:


Where a is the first term, and d is the common difference. Notice that also that the n-th term is given by a + (n – 1)d. This is known as an Arithmetic Progression, or an Arithmetic Sum.

Now, can we find a sum to n terms? That means, can we find a general formula for the following summation of the first term to the last term:


Well, of course we can! Let us see how; first, let us consider the sum to n terms on its own:


We started off by writing the sum from the first term a – but we could have done so by starting from the last term A as well, and thus the sum to n terms written backwards from the last term
A is simply:


Now let’s put them side by side and add them up and let’s see if you notice something:


What do you notice? Well, notice that the ‘d’s and the ‘-d’s all cancel one another out, leaving us with:


The n is there because we have n terms in each sum, and adding up two sums should give us n of A and n of a. Now, what is the last term A in terms of the first term a? We’ve already said that we want the sum to n terms, and thus the last term A must be the n-th term, given by:


In which case we make a substitution:


And therefore the sum to n terms follows nicely:

Sweet eh? :)

Sum To Infinity, and Beyond!

Well, what exactly is a Geometric Progression? To put it very simply, it is a sum or series that has the following pattern:

Of course, you may then ask what is the sum to infinite terms, or the sum to infinity for this series, and we can then represent it as such:

The previous post dealt with the problem of 0.11111… as a recurring decimal using the idea of a Geometric Progression, and I’m about to show you exactly how right now.

First of all, let us consider the sum to n terms (this is often called the partial sum of the first n terms):

It must then be agreeable and logical that if I multiply this by the common ratio r, I obtain:

Taking the difference of the two, I see that:

And rearranging, I immediately see that:

Now what is the sum to infinity then? Easy, it can be evaluated by considering the limit as n tends towards infinity:

Notice that this limit can only exist, if the common ratio r has a magnitude smaller than one, such that it decreases to effectively zero for huge values of n:

With this in mind, the sum to infinity is simply:

So applying this to the previous problem, we see that:

And therefore there is no actual need for any algebraic manipulation to solve for recurring numbers once you have grasped this theory. :)

Simple Algebra, or Not?

Well, I won't be doing much explaining in this post, so if you don't understand maybe I'll post another explanation for the second part; Luoning's tag has directed me to a very nice solution for determining fractional expressions for recurring decimals - it's a method taught to me by Mrs Pauline Kan way back in JC1, and I always turn to it for some mathematical fun when dealing with young Secondary School kids during tuition.

The question goes like this, given a recurring decimal like the one below, can you determine a fraction that equals it:

Well, the trick is to recognise that you can make use of the commutative properties of algebraic manipulation as follows:

And then by equating the two equations we must concur that:

And therefore:

But of course, for those budding Mathematicians out there, you may have seen this as a simple Geometric Progression sum to infinity, and therefore the general formula follows directly for such a sum:

I'm taking for granted that all of you know what Geometric Progressions are, haha.