# Summation - needs to get the constant out

1. May 24, 2013

### liiamra

Hello All,

I have what I think an easy summation, but I haven't worked with math for very long - I don't know the term which I should search the internet for in order to solve the problem and so I would be very thankful if you help me get the constant C out of the summation:

$\sum^{N}_{t=1}$$\beta^{t}$$\frac{C(1-g)^{t}-1}{1-\alpha}$

Where α,β, and C are all constant and I am interested in getting C out.
If it was straight multiplication, I could directly take C out multiplied by Ʃ.
If it was pure addition, I could also take C*N out + Ʃ.

Thanks in Advance,
liiamra

2. May 24, 2013

### liiamra

Just as a quick add:

I think that if I simplify to:

$\sum^{N}_{t=1}$$\frac{\beta^{t}C(1-g)^{t}}{1-\alpha}$-$\frac{\beta^{t}}{1-\alpha}$

I could get the C constant out, but the only problem is that I will end up with:

$C\sum^{N}_{t=1}$$\frac{\beta^{t}(1-g)^{t}}{1-\alpha}$- $\sum^{N}_{t=1}$$\frac{\beta^{t}}{1-\alpha}$

and if the above is correct, the problem is the second term which I want to avoid in order to eliminate C at a later stage by dividing the whole thing by another sum which has the same constant C.

Any ideas

Last edited: May 24, 2013
3. May 25, 2013

### SteamKing

Staff Emeritus
If you multiply each term in a sum by a constant factor C, then you wind up with C * sum. This is basic arithmetic.

4. May 25, 2013

### liiamra

I don't know if I got you right! because:

- I don't want to end up with c*(cƩ-Ʃ) which is nothing really
- Further, 3*2-3*1 are not the same as 2-1

Thanks anyways for the post

5. May 25, 2013

### SteamKing

Staff Emeritus
3*2 - 3*1 = 3*(2 - 1)

This is called the distributive property.

Factoring out a constant from a sum does not mean that the constant disappears.

In the series given in the OP, the denominator, 1 - alpha, is common to all terms of the series and can be factored out.

6. May 25, 2013

### liiamra

Thanks again, but suppose I have the following which can serve as a perfect simplification of the problem:

Ʃa^t*C - a^T

your answer suggests that it becomes ƩC(a^t*C - a^T) and this is not a solution- beside, indeed the denominator can be factored out but I am interested in factoring the C out.

Thanks again and Regards

7. May 25, 2013

### liiamra

Hi again,

The last equation is not equivalent- but the problem, after factoring the (1/1-α) out, can be expressed as:

$\sum^{N}_{t=1}$$\beta^{t}{C(1-g)^{t}-\beta^{t}}$

where I need to get the constant C out.

Regards

8. May 25, 2013

### tiny-tim

welcome to pf!

hello liiamra! welcome to pf!
that's correct
you mean, you have two equations P = C∑A + B, Q = C∑D + E ?

then write C∑A = B - P, C∑D = E - Q, and divide

9. May 25, 2013

### liiamra

Thanks Tin-tim,

No, suppose I have the following two equations:

$x=\sum^{N}_{t=1}$$\beta^{t}{C(1-g)^{t}-\beta^{t}}$
$y=\sum^{N}_{t=1}$$\beta^{t}{C(1+k)^{t}-\beta^{t}}$

I want to calculate $\frac{x}{y}$, and in the process I need to eleminate C as I don't have its value.

Any ideas?

Thanks again and regards//

10. May 25, 2013

### tiny-tim

no, i think the best you can do is find $\frac{x + \beta^t}{y + \beta^t}$

11. May 25, 2013

### liiamra

Thanks Tiny-tim,

The actual equations I am trying to equate are pretty large, and apparently there must be a trick to calculate the variable (found in one of the two equations) - most probably it should be a closed form solution.

So far, I don't think there will be a solution and so I thank you and thank SteamKing for your contributions.

Have a good day//

All Best

12. May 25, 2013

### SteamKing

Staff Emeritus
I have suggested no such thing. The factor must be included in all terms of the series before it can be taken outside the summation. However, you should note the the series in the OP can be re-written as two series, since you have beta^t * C * (1 - g^t) - 1, after factoring out the denominator (1 - alpha).

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