How do you rearrange from $\displaystyle 2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$ to give $\displaystyle 2-\frac{k+3}{2^{k+1}}$
thank you!
Hi BabyMilo!
I think you have made a typing error!
If you mean :
$\displaystyle 2 - \frac{k + 2}{2^k} - \frac{k+1}{2^{k+1}}$
then you can arrange it to get
$\displaystyle 2 - \frac{k+3}{2^{k+1}}$
let me know and if you are still stuck i'll type out the answer!
Lemme know if you need explanation for any particular step.
$\displaystyle 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} = $ $\displaystyle 2- \frac{2(k+2) - (k+1)}{2\times2^k}= 2- \frac{{\color{red}2k+4 - k-1}}{2\times2^k}= 2 - \frac{k+3}{{\color{red}2^{k+1}}}$
$\displaystyle 2-\frac{k+2}{2^k}+\frac{k+1}{2^(k+1)}$
= $\displaystyle 2-\frac{k+2}{2^k}+\frac{k+1}{(2^k) 2}$
= $\displaystyle 2-\frac{2(k+2)}{2^{(k+1)}}+\frac{k+1}{2^{(k+1)}}$
= $\displaystyle 2+\frac{-2k-4+k+1}{2^{(k+1)}}$
= $\displaystyle 2+\frac{-k-3}{2^{(k+1)}}$
= $\displaystyle 2+\frac{-(k+3)}{2^{(k+1)}}$
= $\displaystyle 2-\frac{k+3}{2^{(k+1)}}$
Sorry for the LaTex error on the first post