0

I'm being asked to prove $ 2 \cdot 3^0 + 2 \cdot 3^1 + 2 \cdot 3^2 + \ldots + 2 \cdot 3^{n-1} = 3^n - 1. $ combinatorially using the question "How many length-$n$ lists can we form using the elements in $\{0, 1, 2\}$ in which the elements are not all 0?"

Frankly, I'm not even sure where to start and not entirely sure what a combinatorial proof is, to be honest. How would I go about approaching this problem?

I assume that the right side is $3^n-1$ because it represents the number of lists of length n minus the list of all 0's.

Sanjay B
  • 45

2 Answers2

2

A combinatorial proof shows that an identity is true by demonstrating that the two sides count the same objects in two different ways. For your given identity, you have already determined the objects and an explanation for the RHS. To explain the LHS, condition on the first appearance of $1$ or $2$.

RobPratt
  • 45,619
1

A combinatorial proof is just a proof that depends on counting something. You're correct about the right-hand side.

As to the left hand side, how many lists can you make that don't have a $0$ in the first place? There are $2$ choices for the first place, and $3^{n-1}$ for the remaining places. What about that lists that have a $0$ in the first place but not the second? It's $1\cdot2\cdot3^{n-2}$ and so on.

A neater proof, in my view, is that the left-hand side is $22\dots2_3$ where there are $n$ $2$'s. If we add $1$, the usual addition algorithm gives $10\dots0_3$ where there are $n$ $0$'s.

saulspatz
  • 53,131