number sequences

[bumblebee123]

Pka is starting with the exotic "representative formula" with the scary
"sum symbol" (the one that scares most students!).
The formula you use is correct (same as Pka's):
Sum of 1st n terms = n/2 [2a + ( n - 1 )d ]
Usually shown this way: n[2a + d(n - 1)] / 2

So we have for n terms:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a
= n(2a + 2an - 2a) / 2
= (2an + 2an^2 - 2an) / 2
= 2an^2 / 2
= an^2

Now for 2n terms; we start similarly with:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a and n = 2n:
= 2n(2a + 4an - 2a) / 2
= n(2a + 4an - 2a)
= 2an + 4an^2 - 2an
= 4an^2

It always helps if you make up a simple case; as example:
3, 9, 15 ,21, 27, 33
a = 3
d = 2a = 6
n = 3 (so 2n = 6)

3 + 9 + 15 = 27
3 + 9 + 15 + 21 + 27 + 33 = 108
108 / 27 = 4

would this work if I used my formula? ( as I'm not used to the other one )

a1 = a d = 2a n = N
sN = N/2 [ 2a + ( N -1 ) 2a ] = N/2 ( 2a + 2aN - 2a ) = N/2 ( 2aN ) = N ( aN ) = aN^2

 

[Denis]

would this work if I used my formula? ( as I'm not used to the other one )

Well, get used to it....it's standard and easier...

a1 = a d = 2a n = N
sN = N/2 [ 2a + ( N -1 ) 2a ] = N/2 ( 2a + 2aN - 2a ) = N/2 ( 2aN ) = N ( aN ) = aN^2

That's simply confusing the issue. Plus gives me a headache just looking at it

 

[bumblebee123]

Well, get used to it....it's standard and easier...

That's simply confusing the issue. Plus gives me a headache just looking at it

I've just never been taught this sum by my math teacher. The one that gives you a headache is taught in my math syllabus. It just makes more sense to use something I've been taught and know well, if I was in the exam, to make sure I'd get marks.

I understand the "standard" method though, so I'll make note of it.

 

[Denis]

This is getting more confusing than ever...

In post#20 you said:
I usually use: Sn = n/2 [ 2a + ( n - 1 )d ]

And that's the formula I used.

This thread starts off with this problem:
"An arithmetic series has the first term a and common difference 2a.
Show that the sum to 2N terms is always equal to four times the sum to N terms."

That's a problem your teacher "made up".

 

[Steven G]

Let's get back on track. Using your formula, simplify S_N and S_2N in one post. Is one 4 times the other?