Summation - needs to get the constant out

[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

 

[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

 

[SteamKing]

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

 

[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

 

[SteamKing]

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.

 

[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

 

[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

 

[tiny-tim]

welcome to pf!

hello liiamra! welcome to pf!

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}

that's 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.

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

 

[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//

 

[tiny-tim]

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?

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

 

[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

 

[SteamKing]

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

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).