How Do You Prove the Summation Formula S_{n} = \frac{n}{2}(2a + L)?

[thomas49th]

Prove that
$$S_{n} = \frac{n}{2}(2a + L)$$ where L = a +(n-1)d


Can someone guide me through the proof please

Thx

 

[mathman]

The statement of the problem is confusing. However, the standard proof of summation of arithmetic progression is to take the terms in reverse order and add it to the original sum in direct order, term by term. All pairwise sums are the same, so the sum is the number of terms times one pairwise sum divided by 2.

This is 8th grade or earlier math.

 

[tacman]

for asking for help! Let's go through the proof step by step.

First, let's define the summation formula we want to prove:

S_{n} = a + (a+d) + (a+2d) + ... + (a+(n-1)d)

We can rewrite this as:

S_{n} = (a + (n-1)d) + (a + (n-2)d) + ... + (a + 0d)

Now, let's add this equation to itself, but in reverse order:

S_{n} + S_{n} = (a + (n-1)d) + (a + (n-2)d) + ... + (a + 0d) + (a + 0d) + (a + (n-1)d) + ... + (a + (n-2)d) + (a + (n-1)d)

If we group the terms together, we get:

S_{n} + S_{n} = (2a + (n-1)d) + (2a + (n-1)d) + ... + (2a + (n-1)d)

Since there are n terms in total, we can rewrite this as:

S_{n} + S_{n} = n(2a + (n-1)d)

Now, let's divide both sides by 2 to get the desired formula:

\frac{S_{n} + S_{n}}{2} = \frac{n(2a + (n-1)d)}{2}

Simplifying, we get:

S_{n} = \frac{n}{2}(2a + (n-1)d)

But we can also rewrite L = a + (n-1)d as:

L = a + (n-1)d = a + (n-1)d

Substituting this into our formula, we get:

S_{n} = \frac{n}{2}(2a + L)

And there you have it, we have proven the summation formula! I hope this helps guide you through the proof. Let me know if you have any further questions.