(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

All I am trying to do is the basic mathematical induction routine: n=1, n=k, n=k+1 and how n=k proves n=k+1. The problem I am having is with the algebra.

[tex]4+16+64+...+4^{n}=\frac{4}{3}(4^{n}-1)[/tex]

2. Relevant equations and Attempts

For n=1

[tex]4=4[/tex] That's OK

For n=k

[tex]4+16+64+...+4^{k}=\frac{4}{3}(4^{k}-1)[/tex]

For n=k+1 by Substitution

[tex]4+16+64+...+4^{(k+1)}=\frac{4}{3}(4^{(k+1)})[/tex]

To Prove by Adding [tex]K+1[/tex] to [tex]S_{k}[/tex]

[tex]4+16+64+...+4^{k}+4^{(k+1)}=\frac{4}{3}(4^{k}-1)+4^{(k+1)}[/tex]

It took me a minute to figure out how [tex]\frac{4}{3}(4^{k}-1)[/tex] became [tex]\frac{4^{(k+1)}}{3}[/tex], but I got that part.

[tex]= \frac{4^{(k+1)}}{3}+4^{(k+1)}-\frac{4}{3}[/tex]

This is where I am stuck. I gather that I am supposed to multiply [tex]4^{(k+1)}[/tex] by [tex]\frac{3}{3}[/tex], and then combine like terms. I am confused about this because I don't know how to multiply the term by 3 (I probably learned it, I'm just stumped).

Please help me understand what to do next. As long as I can understand how to manipulate the exponential terms with the fractions, I can complete this. Thanks for your time.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Basic Mathematical Induction w/ n In the Exponent

**Physics Forums | Science Articles, Homework Help, Discussion**