Jameson Lucero

Answered

2022-07-07

algebra exponents and fractions

I could be over thinking or tired... But I am to embarrassed to ask my prof. this probably very simple algebra rule I am ignorant of... Also this is just a snip-it from a inductive proof example.

Say you have something like this

$\frac{{4}^{k}-1}{3}+\frac{3\ast {4}^{k}}{3}=$

$\frac{4\ast {4}^{k}-1}{3}$

My question is what algebra rules or rules of fractions allow this? In other words what computation is going on? I can agree that the 3 should cancel out

$\frac{3\ast {4}^{k}}{3}={4}^{k}$

But wouldn't that mean it be something like this

$\frac{({4}^{k}-1)\ast {4}^{k}}{3}$

or more accurately

$\frac{({4}^{k}-1)}{3}\ast {4}^{k}$

and not this?

$\frac{4\ast {4}^{k}-1}{3}$

I could be over thinking or tired... But I am to embarrassed to ask my prof. this probably very simple algebra rule I am ignorant of... Also this is just a snip-it from a inductive proof example.

Say you have something like this

$\frac{{4}^{k}-1}{3}+\frac{3\ast {4}^{k}}{3}=$

$\frac{4\ast {4}^{k}-1}{3}$

My question is what algebra rules or rules of fractions allow this? In other words what computation is going on? I can agree that the 3 should cancel out

$\frac{3\ast {4}^{k}}{3}={4}^{k}$

But wouldn't that mean it be something like this

$\frac{({4}^{k}-1)\ast {4}^{k}}{3}$

or more accurately

$\frac{({4}^{k}-1)}{3}\ast {4}^{k}$

and not this?

$\frac{4\ast {4}^{k}-1}{3}$

Answer & Explanation

jugf5

Expert

2022-07-08Added 18 answers

$\frac{{4}^{k}-1}{3}+\frac{3\cdot {4}^{k}}{3}\phantom{\rule{0ex}{0ex}}\text{these fractions have the same denominator so we can write the fraction as one now}\phantom{\rule{0ex}{0ex}}\frac{{4}^{k}-1+3\cdot {4}^{k}}{3}\phantom{\rule{0ex}{0ex}}\text{recall}u+3u=4u\phantom{\rule{0ex}{0ex}}\text{so we have}\phantom{\rule{0ex}{0ex}}\frac{{4}^{k}(1+3)-1}{3}=\frac{{4}^{k}(4)-1}{3}=\frac{4\cdot {4}^{k}-1}{3}=\frac{{4}^{k+1}-1}{3}\phantom{\rule{0ex}{0ex}}$

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