Describe your thinking and the steps taken when developing an
algebraic representation for the sum of the first t terms of a
linear sequence.

Let's see what to do buddy...

_________________________________

Step (1)

Suppose we want to find the sum of numbers from 1 to n.

To do this, we use the following method.

By writing the sum of numbers from 1 to n

, Then we write the sum of numbers n to 1 below it, see :

[tex]s = 1 + 2 + 3 + ... + n[/tex]

[tex]s = n + (n - 1) + (n - 2) + ... + 1 \\ [/tex]

Now we add the sentences of the above two phrases, peer to peer like this :

[tex]1 + n = n + 1 \\ [/tex]

[tex]2 + n - 1 = n + 1[/tex]

[tex]3 + n - 2 = n + 1[/tex]

And the others.....

So we have :

[tex]2s = ( n + 1) + (n + 1) + ... + (n + 1) \\ [/tex]

We had n numbers to sum so we have :

[tex]2s = n \times (n + 1)[/tex]

Divided the sides of the equation by 2

[tex]s = \frac{n \times (n + 1)}{2} \\ [/tex]

Remember this step I will use it again.

_________________________________

Step (2)

What does the linear sequence mean ?

The linear sequence is the sequence which any terms created by the sum of previous term with constant.

I name that constant value d.

According to above :

[tex]t(2) = t(1) + d[/tex]

And

[tex]t(3) = t(2) + d [/tex]

[tex]t(3) = t(1) + d + d[/tex]

[tex]t(3) = t(1) + 2d[/tex]

I have a question ;

[tex]3 - 1 = 2[/tex]

Is it correct ?

If it is correct we have :

[tex]t(3) = t(1) + (3 - 1)d[/tex]

WOW we found a thing ;

Put n instead of 3 :

[tex]t(n) = t(1) + (n - 1)d[/tex]

_________________________________

Step (3)

Stop right here.

Let's go to find the sum of the n first terms of the linear sequence.

Do you remember what did we do in step(1) ? Of course you do.

Let's do it again.

[tex]s = t(1) + t(2) + t(3) + ... + t(n) \\ [/tex]

[tex]s = t(n) + t(n - 1) + t(n - 2) + ... + t(2) + t(1) \\ [/tex]

According to the thing what we found in step(2) we have :

[tex]s = t(1) + ( \: t(1) + d \: ) + ( \: t(1) + 2d \: ) + ... + ( \: t(1) + (n - 1)d \: ) \\ [/tex]

[tex]s = ( \: t(1) + (n - 1)d \: ) + ( \: t(1) + (n - 2)d \: ) + ... + t(1) \\ [/tex]

Sum the two above equation's terms like this:

[tex]t(1) + t(1) + (n - 1)d = 2t(1) + (n - 1)d \\ [/tex]

And

[tex]t(1) + d +t(1) + (n - 2)d = 2t(1) + d(n - 2 + 1) = 2t(1) + (n - 1)d \\ [/tex]

And the others like this.

We had n terms so we sumed n terms.

So we have :

[tex]2s = n \times ( \: 2t(1) + (n - 1)d \: )[/tex]

Divided the sides of the equation by 2

[tex]s(n) = \frac{n}{2} \times ( \: 2t(1) + (n - 1)d \: ) \\ [/tex]

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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