# Thread: Simplifying Expressions

1. ## Simplifying Expressions

Hello,

I wasn't sure if this belongs in Pre-Calc, Algebra, or elsewhere, so I hope it doesn't get deleted.

I have a couple of expressions I'm trying to simplify and either I'm erring in the simplification, or I am not boiling them down to their "simplest" forms.

I have $\displaystyle a_k = \frac{2(-1)^k2^k}{k!}$ and I need to find $\displaystyle |\frac{a_{k+1}}{a_k}|$

So,

$\displaystyle \frac{\frac{2(1)^{k+1}(2)^{k+1}}{(k+1)!}}{\frac{2( 1)^k(2)^k}{k!}}$ $\displaystyle = \frac{2(1)^{k+1}(2)^{k+1}k!}{2(1)^k(2)^k(k+1)!}$ = $\displaystyle \frac{2^{k+1}k!}{2^k(k+1)!}$ which is not recognized as correct by the system.

There is also another expression $\displaystyle b_k = \frac{2(-1)^kcos(k\pi)}{k}$ for which I need to find $\displaystyle |\frac{b_{k+1}}{b_k}|$

So,

$\displaystyle \frac{\frac{2(1)^{k+1}cos((k+1)\pi)}{k+1}}{\frac{2 (1)^kcos(k\pi)}{k}}$ $\displaystyle = \frac{2(1)^{k+1}cos((k+1)\pi)k}{2(1)^kcos(k\pi)(k+ 1)}$ $\displaystyle = \frac{cos((k+1)\pi)k}{cos(k\pi)(k+1)}$

I'm not sure where I'm faltering. If anyone could point out my mistakes, I would be grateful. Again, sorry if this is in the wrong section.

Thanks for your help,

Austin

2. Originally Posted by auslmar
Hello,

I wasn't sure if this belongs in Pre-Calc, Algebra, or elsewhere, so I hope it doesn't get deleted.

I have a couple of expressions I'm trying to simplify and either I'm erring in the simplification, or I am not boiling them down to their "simplest" forms.

I have $\displaystyle a_k = \frac{2(-1)^k2^k}{k!}$ and I need to find $\displaystyle |\frac{a_{k+1}}{a_k}|$

So,

$\displaystyle \frac{\frac{2(1)^{k+1}(2)^{k+1}}{(k+1)!}}{\frac{2( 1)^k(2)^k}{k!}}$ $\displaystyle = \frac{2(1)^{k+1}(2)^{k+1}k!}{2(1)^k(2)^k(k+1)!}$ = $\displaystyle \frac{2^{k+1}k!}{2^k(k+1)!}$ which is not recognized as correct by the system.

There is also another expression $\displaystyle b_k = \frac{2(-1)^kcos(k\pi)}{k}$ for which I need to find $\displaystyle |\frac{b_{k+1}}{b_k}|$

So,

$\displaystyle \frac{\frac{2(1)^{k+1}cos((k+1)\pi)}{k+1}}{\frac{2 (1)^kcos(k\pi)}{k}}$ $\displaystyle = \frac{2(1)^{k+1}cos((k+1)\pi)k}{2(1)^kcos(k\pi)(k+ 1)}$ $\displaystyle = \frac{cos((k+1)\pi)k}{cos(k\pi)(k+1)}$

I'm not sure where I'm faltering. If anyone could point out my mistakes, I would be grateful. Again, sorry if this is in the wrong section.

Thanks for your help,

Austin
The problem might simply be that you can simplify each of these still more:
$\displaystyle \frac{2^{k+1}k!}{2^k(k+1)!} = \frac{2}{k + 1}$

and
$\displaystyle \frac{cos((k+1)\pi)k}{cos(k\pi)(k+1)} = -\frac{k}{k + 1}$

-Dan