>I think this would work mo' bettah.
>
[quoted text clipped - 5 lines]
>
>Gord Dibben MS Excel MVP
Well Gord, with no intention of implying that I'm well versed in
mathematics, I just happen to take a liking to this "Arithmetic Series".
A year or so ago, Dana put me on to a link (which I can't find at the
moment) that described how Gauss formulated this "truth".
In words it's:
<<<"The count of sequential numbers times the average of the first and last
number.">>>
=A1*(A1+1)/2
So, the formula I posted is supposed to be "intuitive" for *ALL*
possibilities.
Here, the OP's request was for a sum of a series starting (or ending) with
1, therefore, the last (or first) number in this series is the *actual*
amount (count) of numbers in the sequential series, so your A1^2 works.
BUT ... what if the we're looking for, say 5 to 15?
A1 = 15
B1 = 5
In the original formula, A1 now has to become (A1-B1+1)
And the balance becomes (A1+B1)/2
So we now have:
=(A1-B1+1)*(A1+B1)/2
Taking it a step further, there's *no* stipulation that the sequential
series must have an increment or decrement constant of *1*.
Let's take the same 5 to 15, but here we want an increment constant of 5,
the SUM of the sequential series of 5, 10, &15.
A1 = 15
B1 = 5
C1 = 5
=((A1-B1)/C1+1)*(A1+B1)/2
So, that's why:
<<<"The count of sequential numbers times the average of the first and last
number.">>>
=A1*(A1+1)/2
Is probably easier to relate to.

Signature
Regards,
RD
-----------------------------------------------------------------------------------------------
Please keep all correspondence within the Group, so all may benefit !
-----------------------------------------------------------------------------------------------
oops!!
Not better, just the same.
Sorry RD
>I think this would work mo' bettah.
>
[quoted text clipped - 8 lines]
>
>Gord Dibben MS Excel MVP
Gord Dibben MS Excel MVP
Gord Dibben - 21 Oct 2006 21:16 GMT
RD
I will ponder this for a while then quietly drift off to whatever planet I
currently reside on. Depends upon which set of meds I'm on<g>
But I am improving.....honestly!
Thanks for the work you put into this explanation.
Saved for future regurgitation.
Gord
>Well Gord, with no intention of implying that I'm well versed in
>mathematics, I just happen to take a liking to this "Arithmetic Series".
[quoted text clipped - 43 lines]
>
>Is probably easier to relate to.
Gord Dibben MS Excel MVP
RagDyeR - 21 Oct 2006 22:21 GMT
Found that link!
Here it is:
http://mathworld.wolfram.com/ArithmeticSeries.html

Signature
Regards,
RD
-----------------------------------------------------------------------------------------------
Please keep all correspondence within the Group, so all may benefit !
-----------------------------------------------------------------------------------------------
RD
I will ponder this for a while then quietly drift off to whatever planet I
currently reside on. Depends upon which set of meds I'm on<g>
But I am improving.....honestly!
Thanks for the work you put into this explanation.
Saved for future regurgitation.
Gord
>Well Gord, with no intention of implying that I'm well versed in
>mathematics, I just happen to take a liking to this "Arithmetic Series".
[quoted text clipped - 43 lines]
>
>Is probably easier to relate to.
Gord Dibben MS Excel MVP