Basic Mathematical Induction w/ n In the Exponent

[DDNow]

Homework Statement

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.

$$4+16+64+...+4^{n}=\frac{4}{3}(4^{n}-1)$$

2. Homework Equations and Attempts

For n=1

$$4=4$$ That's OK

For n=k

$$4+16+64+...+4^{k}=\frac{4}{3}(4^{k}-1)$$

For n=k+1 by Substitution

$$4+16+64+...+4^{(k+1)}=\frac{4}{3}(4^{(k+1)})$$

To Prove by Adding $$K+1$$ to $$S_{k}$$

$$4+16+64+...+4^{k}+4^{(k+1)}=\frac{4}{3}(4^{k}-1)+4^{(k+1)}$$

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

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

This is where I am stuck. I gather that I am supposed to multiply $$4^{(k+1)}$$ by $$\frac{3}{3}$$, 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.

 

[Vagrant]

For n=k+1 by Substitution

$$4+16+64+...+4^{(k+1)}=\frac{4}{3}(4^{(k+1)})$$

Check this expression.
Once you have the correct target expression, the next part should work out.

 

[gabbagabbahey]

ummm...when you multiply $$4^{k+1}$$ by $$\frac{3}{3}$$ you just get $$\frac{3(4^{k+1})}{3}$$...and since you now have two terms with the same denominator, you can add the numerators to get: $$\frac{4^{(k+1)}+3(4^{(k+1)})}{3}-\frac{4}{3}$$

what do you get when you add one of 'something' to 3 of 'something'?

 

[DDNow]

Check this expression.
Once you have the correct target expression, the next part should work out.

I'm not sure how that is incorrect.

ummm...when you multiply $$4^{k+1}$$ by $$\frac{3}{3}$$ you just get $$\frac{3(4^{k+1})}{3}$$...and since you now have two terms with the same denominator, you can add the numerators to get: $$\frac{4^{(k+1)}+3(4^{(k+1)})}{3}-\frac{4}{3}$$

what do you get when you add one of 'something' to 3 of 'something'?

I get it now. When I look at it like 1[4^(K+1)]+3[4^(K+1)], it makes sense. I was trying to multiply, which was wrong, and didn't let me pull out 4/3.

Thanks, everyone.

 

[Vagrant]

When you substitute n=k+1,
right hand side of the expression should be $$4/3$$($$4^{k+1}$$ -1)