Rearranging Algebra

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March 19th 2010, 11:29 AM
BabyMilo

Rearranging Algebra

How do you rearrange from $2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$ to give $2-\frac{k+3}{2^{k+1}}$

thank you!
March 19th 2010, 11:50 AM
s_ingram

Hi BabyMilo!

I think you have made a typing error!

If you mean :

$2 - \frac{k + 2}{2^k} - \frac{k+1}{2^{k+1}}$

then you can arrange it to get

$2 - \frac{k+3}{2^{k+1}}$

let me know and if you are still stuck i'll type out the answer!
March 19th 2010, 11:50 AM
Anonymous1

Lemme know if you need explanation for any particular step.

$2- \frac{k+2}{2^k} - \frac{k+1}{2^{k+1}} = 2- \frac{k+2}{2^k} - \frac{k+1}{{\color{red} 2\times2^k}} = 2- \frac{{\color{red} 2}(k+2)}{{\color{red} 2}\times2^k} - \frac{k+1}{2\times2^k} =$ $2- \frac{2(k+2) - (k+1)}{2\times2^k}= 2- \frac{{\color{red}2k+4 - k-1}}{2\times2^k}= 2 - \frac{k+3}{{\color{red}2^{k+1}}}$
March 19th 2010, 11:51 AM
harish21

Quote:

Originally Posted by BabyMilo

How do you rearrange from $2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$ to give $2-\frac{k+3}{2^{k+1}}$

thank you!

$2-\frac{k+2}{2^k}+\frac{k+1}{2^(k+1)}$

= $2-\frac{k+2}{2^k}+\frac{k+1}{(2^k) 2}$

= $2-\frac{2(k+2)}{2^{(k+1)}}+\frac{k+1}{2^{(k+1)}}$

= $2+\frac{-2k-4+k+1}{2^{(k+1)}}$

= $2+\frac{-k-3}{2^{(k+1)}}$

= $2+\frac{-(k+3)}{2^{(k+1)}}$

= $2-\frac{k+3}{2^{(k+1)}}$

Sorry for the LaTex error on the first post
March 19th 2010, 12:04 PM
BabyMilo

Quote:

Originally Posted by harish21

$2-\frac{k+2}{2^k}+\frac{k+1}{2^(k+1)}$

= $2-\frac{k+2}{2^k}+\frac{k+1}{(2^k) 2}$

= $2-\frac{2(k+2)}{2^{(k+1)}}+\frac{k+1}{2^{(k+1)}}$

= $2+\frac{-2k-4+k+1}{2^{(k+1)}}$

= $2+\frac{-k-3}{2^{(k+1)}}$

= $2+\frac{-(k+3)}{2^{(k+1)}}$

= $2-\frac{k+3}{2^{(k+1)}}$

Sorry for the LaTex error on the first post

the steps made it very clear.

thanks

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