Inductive proof foundations of analysis

[inovermyhed]

Homework Statement

Prove the following. 4+1+4^-1+4^-2...4^-k = 16/3-1/(3)(4^k)

Homework Equations

The Attempt at a Solution

I confirmed the basis step and proceeded by changing the variable to n and adding n+1. I now have (16/3) - (1/(3)(4^k)) + (1/4^k+1) = 16/3 - (1/(3)(4^k+1)). In trying to get these even I have tried simplifying (that made a mess), adding and subtracting by (3)(12k), multiplying and dividing by 4^k+1 and a combination of these tactics. I just need some direction on how to proceed or to know if I am going about this completely wrong. Thank you.

 

[SammyS]

Homework Statement

Prove the following. 4+1+4^-1+4^-2...4^-k = 16/3-1/(3)(4^k)

Homework Equations

The Attempt at a Solution

I confirmed the basis step and proceeded by changing the variable to n and adding n+1. I now have (16/3) - (1/(3)(4^k)) + (1/4^k+1) = 16/3 - (1/(3)(4^k+1)). In trying to get these even I have tried simplifying (that made a mess), adding and subtracting by (3)(12k), multiplying and dividing by 4^k+1 and a combination of these tactics. I just need some direction on how to proceed or to know if I am going about this completely wrong. Thank you.

Hello inovermyhed. Welcome to PF !

Making your post easier to read, will likely get you more help, faster.

Your post would be much easier to read, if you would use appropriate spacing --- even better yet, in addition to that, use the X² & X₂ icons for super/sub scripts --- even better yet use LaTeX.

I.E.:

[ QUOTE ]

Homework Statement

Prove the following. 4+1+4^-1+4^-2...4^-k = 16/3-1/(3)(4^k)[ /QUOTE ]​

I suppose the (4^k) should actually be in the denominator. ?

[ QUOTE ]

Homework Equations

The Attempt at a Solution

I confirmed the basis step and proceeded by changing the variable to n and adding n+1.

I now have

(16/3) - (1/(3)(4^k)) + (1/4^k+1) = 16/3 - (1/(3)(4^k+1)).​

In trying to get these even I have tried simplifying (that made a mess), adding and subtracting by (3)(12k), multiplying and dividing by 4^k+1 and a combination of these tactics. I just need some direction on how to proceed or to know if I am going about this completely wrong. Thank you.
[/QUOTE ]​

"QUOTE" tags intentionally deactivated so OP can see what's behind the spacing & formatting.

 

[dot.hack]

This is coming from another Real Analysis student so I may be making a mistake, but the practice for me is definitely worth it so please point out if I did something wrong.
Taking k=1 we see that the equality holds, so we assume the equality holds for k and attempt to prove for k+1. We then have ${\frac{16}{3}} - {\frac{1}{3 * 4^k}} + 4^{-k -1} = {\frac{16}{3}} - {\frac{1}{3 * 4^{k+1}}}$
Subtract 16/3 from both sides and multiply by negative 3 to get
${\frac{1}{{4^k}}} - 3 * 4^{-k -1} = {\frac{1}{4^{k+1}}}$
You can then multiply both sides by $4^{k+1}$ to get
${\frac{4^{k+1}}{4^k}} - 3*4^{k+1} * 4^{-k -1} = 1$
but the first term on the left just gives 4 and through a little manipulation i.e breaking $4^{k+1}$ into $4^k *4$ and a similar maneuver with $4^{-k -1}$ you get 4-3 = 1 thus proving the equality.
I think that is correct.