# Thread: Problems solving an equation with sum

1. ## Problems solving an equation with sum

I'd like to solve this equation with respect to q.

$\displaystyle +(100000 - 2500) - \Sigma_{t=1}^{48}\frac{2500 + 60}{(1+q)^{t}} = 0$

$\displaystyle 97500 = \Sigma_{t=1}^{48}\frac{2560}{(1+q)^{t}}$

Am I allowed to multiply by $\displaystyle (1+q)^{t}$ on both sides of the equation?

I don't understand how I can isolate q without doing so...

$\displaystyle q = \Sigma_{t=1}^{48}\sqrt[t]{\frac{2560}{97500}} - 1$

I'm lost here, any help would be very much appreciated.

2. Originally Posted by MatteNoob
I'd like to solve this equation with respect to q.

$\displaystyle +(100000 - 2500) - \Sigma_{t=1}^{48}\frac{2500 + 60}{(1+q)^{t}} = 0$

$\displaystyle 97500 = \Sigma_{t=1}^{48}\frac{2560}{(1+q)^{t}}$

Am I allowed to multiply by $\displaystyle (1+q)^{t}$ on both sides of the equation?

I don't understand how I can isolate q without doing so...

$\displaystyle q = \Sigma_{t=1}^{48}\sqrt[t]{\frac{2560}{97500}} - 1$

I'm lost here, any help would be very much appreciated.
Try writing the sum (which is a Geometric Series) in its closed form...

3. Ok, let me try again, I'll do some substitutions for

$\displaystyle +(100000 - 2500) - \Sigma_{t=1}^{48}\frac{2500 + 60}{(1+q)^{t}} = 0$

$\displaystyle +(G) - \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}} = 0$

so

$\displaystyle G = \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}}$

$\displaystyle G = \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}}$

$\displaystyle G \cdot \left(\frac{F}{(1+q)} - 1\right) = \left(\frac{F}{(1+q)}\right)^{x} - 1$

$\displaystyle \frac{GF}{(1+q)} - G + 1 = \left(\frac{F}{(1+q)}\right)^{x}$

I take my chances.

$\displaystyle \left(\frac{F}{(1+q)}\right)^{x} - \frac{GF}{(1+q)} = 1 - G$

Sorry I have no clue, can someone help me?

4. Originally Posted by MatteNoob
Ok, let me try again, I'll do some substitutions for

$\displaystyle +(100000 - 2500) - \Sigma_{t=1}^{48}\frac{2500 + 60}{(1+q)^{t}} = 0$

$\displaystyle +(G) - \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}} = 0$

so

$\displaystyle G = \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}}$

$\displaystyle G = \Sigma_{t=1}^{x}\frac{F}{(1+q)^{t}}$

$\displaystyle G \cdot \left(\frac{F}{(1+q)} - 1\right) = \left(\frac{F}{(1+q)}\right)^{x} - 1$

$\displaystyle \frac{GF}{(1+q)} - G + 1 = \left(\frac{F}{(1+q)}\right)^{x}$
I have no idea how you got this or where that "x" came from. There was no "x" in the problem before!

When Prove It said "write the sum in its closed form" he meant that the sum of a geometric series: $\displaystyle \sum_{i=1}^n ar^i$ can be written in the form $\displaystyle \frac{ar(1- r^{n+1}}{1- r}$. Here, a= F and $\displaystyle r= \frac{1}{1+q}$ so the sum is $\displaystyle \frac{F(1- \frac{1}{(1+q)^{49}})}{(1+ q)q}$$\displaystyle = \frac{F((1+q)^{49}- 1)}{q(1+q)^{50}}$.

Your equation is $\displaystyle G= \frac{F((1+q)^{49}- 1)}{q(1+q)^{50}}$ or, equivalently, $\displaystyle q(1+q)^{50}G= f((1+q)^{49}- 1)$.
I take my chances.

$\displaystyle \left(\frac{F}{(1+q)}\right)^{x} - \frac{GF}{(1+q)} = 1 - G$

Sorry I have no clue, can someone help me?