Prove that we can get any number by just adding and subtracting distinct exponents of 3

Question

It is a proof for the solution of the puzzle here.

Answer 1

We need to prove that for any $n$ ,we can write a number by just adding or subtracting the exponents of $3$ till $3^n$

Proof by induction

We know that $$ 3^0 + 3^1 + 3^2 + … 3^n = \frac {3^{n+1} - 1}{3-1}$$ Hence $$3^{n+1} - (3^0 + 3^1 + 3^2 + … 3^n) = 3^0 + 3^1 + 3^2 + … 3^n + 1$$ We already know that we can write every number $(1,2,3…..)$ till $(3^0 + 3^1 + 3^2 + … 3^n,)$ hence we can write any number from $(3^0 + 3^1 + 3^2 + … 3^n + 1)$ till $(3^{n+1} - 1)$ by plugging in the equation for a number k in the set $$(1,2….., 3^0 + 3^1 + 3^2 + … 3^n )$$ in the equation $$3^{n+1} - k $$ After which we will need to add k, that is $$3^{n+1} + k$$ to get every number till $$3^0 + 3^1 + 3^2 + … 3^n + 3^{n+1}$$ For the induction to be complete we will need to prove it for the case of n = 1, which is fairly simple As $(1 = 3^0, 2 = 3^1 - 3^0, 3 = 3^1)$ And the induction is complete !!!