0

I'm having trouble combinatorially proving: $$2 \cdot 3^0 + 2 \cdot 3^1 + 2 \cdot 3^2 + \cdots 2 \cdot 3^{n-1} = 3^n-1$$

I understand that the right hand side can be expressed as the number of possible n-length strings from an alphabet {a,b,c}, minus some specific string, such as "abc".

I see also that the left hand side is a finite geometric series where a = 2 and r = 3 and that I can algebraically prove its sum to be $3^n - 1$.

But how can I prove this combinatorially? How can I show the geometric series to be an answer to the same n-length string problem as the right side, if that is the correct interpretation for the proof?

rajrishidas
  • 11
  • Welcome to Mathematics Stack Exchange. As I commented to this question, $ 2+6+18+⋯+2⋅3n−1=(3−1)+(3−1)⋅3+(3−1)⋅3^2+⋯+(3−1)⋅3^{n−1} $ telescopes to $3⋅3^n−1−1=3^n−1$ – J. W. Tanner Feb 24 '20 at 23:11
  • 1
    OP specifically wants a combinatorial interpretation of the left hand side – TheEmptyFunction Feb 24 '20 at 23:14
  • 2
    Note that $3^n-1$ is the number of all ternary strings (strings made from symbols $0$, $1$, and $2$) of length $n$ that is not $000\ldots0$. The number of such strings such that $k$ is the last (rightmost) non-zero location is given by $2\cdot 3^{k-1}$. – Batominovski Feb 24 '20 at 23:16

1 Answers1

0

$3^n-1$ is the number of strings of length $n$ with characters $\{a,b,c\}$ that isn't $\underbrace{aa\ldots a}_n$.

If the first character is $b,c$, the remainder of the characters can be whatever, so there are $2\cdot3^{n-1}$ strings of this form.

If the first character is $a$, then the number of strings of length $n$ with characters from $\{a,b,c\}$ not including $\underbrace{aa\ldots a}_n$ is the same as the number of strings of length $n-1$ with the same properties, of which there are $3^{n-1}-1$.

So, combinatorially, you can see that $3^n-1=2\cdot3^{n-1}+3^{n-1}-1$, which is the desired relation.

Rushabh Mehta
  • 13,663