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I've just started a course in mathematics at university, and our current topic is mathematical induction.

I've been given the following question:

$$1+4+4^2+....+4^{n-1}=\frac{4^{n}-1}{3}.$$

I get the first step - $P(1) = 1$.

I get the second step - Assume $n = k$.

It's this third step that gets me. I've seen it done a few ways.. but this is what I've got so far:

$$\begin{align} 1+4+4^2+...+4^{k-1}+4^{k} & = [1+4+4^2+...+4^{k-1}] +4^{k} \\ & = \frac{4^{k}-1}{3} + 4^{k} \end{align}$$

I've no idea where to go after this. I'm assuming I want it to look something like:

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

However, no idea how to get there, or if that's even the direction I want to be heading in.

I would appreciated any help. Especially if there's an easier way of doing this.

Cheers. =)

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  • $\begingroup$ try looking up: Sum of terms in geometric progression; see the "sum" subsection here en.wikipedia.org/wiki/Geometric_series. for an explanation. Goliath Madison =user 99680 $\endgroup$ – user99680 Jul 5 '14 at 8:36
  • $\begingroup$ Welcome to Math SE! Using LaTeX would make your formulae clear. Also, your formula for the sum is wrong, so I don't know how you proved the base case. $\endgroup$ – user21820 Jul 5 '14 at 8:47
  • $\begingroup$ @Fantini. I am afraid that the edit is wrong. Cheers. $\endgroup$ – Claude Leibovici Jul 5 '14 at 8:53
  • $\begingroup$ @ClaudeLeibovici I've merely added LaTeX Claude, can you point where I possibly interpreted a formula wrong? Cheers. =) $\endgroup$ – Mark Fantini Jul 5 '14 at 8:54
  • $\begingroup$ @Fantini. Have a look to the first line of my answer. This was in the post. The remaining just follows. Sorry for not doing it myself but, being almost blind, this is very tiring to me. Cheers. $\endgroup$ – Claude Leibovici Jul 5 '14 at 8:58
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You have $$\sum_{k=0}^{n-1}4^k=\frac{1}{3} \left(4^n-1\right)$$ Let us go for the next term; so $$\sum_{k=0}^{n}4^k=\sum_{k=0}^{n-1}4^k+4^n=\frac{1}{3} \left(4^n-1\right)+4^n=\frac{1}{3} (4^n-1+3 \times4^n)$$ $$=\frac{1}{3} (4^n-1+(4-1) \times4^n)=\frac{1}{3} (4^n-1+ 4^{n+1}-4^n)=\frac{1}{3} (4^{n+1}-1)$$

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  • $\begingroup$ @Fantini. It seems to be correct. Thank you very much. I appreciate your good will, be sure. $\endgroup$ – Claude Leibovici Jul 5 '14 at 9:00

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