# Simplifying Expressions

• Jul 6th 2008, 04:20 PM
auslmar
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.

Austin
• Jul 6th 2008, 05:23 PM
topsquark
Quote:

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.

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