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. =)
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 $$\sum_{k=0}^{n}4^k=\sum_{k=0}^{n-1}4^k+4^n=\frac{1}{3} \left(4^n-1\right)+4^n=\frac{1}{3} (4^n-1+3 \times4^n)$$ $$=\frac{1}{3} (4^n-1+(4-1) \times4^n)=\frac{1}{3} (4^n-1+ 4^{n+1}-4^n)=\frac{1}{3} (4^{n+1}-1)$$