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.
♥️♥️♥️♥️♥️
