<s>[INST] Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
 [/INST] Forget about all the smart generic formulas. Just rewrite the last summand $1$ as $2-1$. You get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2 + 2 - 1.
$$
Now group the $2$'s together. $2 + 2 = 2^2$, so you get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2^2 - 1
$$
Now group $2^2$ and $2^2$ together to get $2^2 + 2^2 = 2^3$:
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^3 - 1
$$
If you go on in this fashion, you will eventually come to $2^n - 1$. The actual number of steps in this process depends on $n$, so technically this should be framed as a proof by induction.
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