1. How would one go about solving this equation for a:

2.

3. I'd start by finding a closed form for the sum (one exists), and go from there.

4. Originally Posted by MagiMaster
I'd start by finding a closed form for the sum (one exists), and go from there.
In terms of , but that would be rather tedious.

5. Check out the Wikipedia article http://en.wikipedia.org/wiki/Arithmetic_series.

6. Originally Posted by Ellatha
How would one go about solving this equation for a:

This formula can be proved by induction or by the following trick
Code:
1              2            3     ...     N
N             N-1         N-2  .....      1
--------------------------------------------
(N+1) +     (N+1) +      (N+1)  + ...+ (N+1)
You should be able to use this to solve your problem.

7. Originally Posted by DrRocket
Originally Posted by Ellatha
How would one go about solving this equation for a:

This formula can be proved by induction or by the following trick
Code:
1              2            3     ...     N
N             N-1         N-2  .....      1
--------------------------------------------
(N+1) +     (N+1) +      (N+1)  + ...+ (N+1)
You should be able to use this to solve your problem.
But that is for the case that , but for my formula, it is not , but . Also, n does not equal 1, but 99. Would I need to manipulate the formula?

8. Yes. Try a verbal description of the formula. The series 2+3+4+5 is the same as (5+2)+(3+4) [associative and commutative], however, the second representation has the advantage that each term is the same (i.e. 7). This can be done with the sum of any series of consecutive numbers. Changing n to n+1 just shifts where the series begins and ends. If you are adding X numbers, then there are X/2 pairs of numbers with the same value. In your formula 100 is the first number in the series (99+1) and a+1 is the last number in the series. So all you have to do is figure out how many pairs of number there are in the series, and what the value of each of the pairs are (which is simply the first number in the series plus the last number in the series) and you are good to go.

9. Originally Posted by Ellatha
Originally Posted by DrRocket
Originally Posted by Ellatha
How would one go about solving this equation for a:

This formula can be proved by induction or by the following trick
Code:
1              2            3     ...     N
N             N-1         N-2  .....      1
--------------------------------------------
(N+1) +     (N+1) +      (N+1)  + ...+ (N+1)
You should be able to use this to solve your problem.
But that is for the case that , but for my formula, it is not , but . Also, n does not equal 1, but 99. Would I need to manipulate the formula?
You you still need to a little bit of work. But it should be pretty easy.

For instance

and

10. Originally Posted by DrRocket
Originally Posted by Ellatha
Originally Posted by DrRocket
Originally Posted by Ellatha
How would one go about solving this equation for a:

This formula can be proved by induction or by the following trick
Code:
1              2            3     ...     N
N             N-1         N-2  .....      1
--------------------------------------------
(N+1) +     (N+1) +      (N+1)  + ...+ (N+1)
You should be able to use this to solve your problem.
But that is for the case that , but for my formula, it is not , but . Also, n does not equal 1, but 99. Would I need to manipulate the formula?
You you still need to a little bit of work. But it should be pretty easy.

For instance

and

Alright, here is my work:

11. Originally Posted by Ellatha

Alright, here is my work:

Close

Should be

Which still gets you

But is supposed to be an integer, so there really is no exact solution.

You can calculate that

and

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