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? :)
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