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$$\sum_{i=0}^j {j \choose i}2^{j-i} = 3^j$$

My approach: I know the binomial way to do this is to think of the RHS as $(1+2)^j$ and then expand using binomial like so: $$(1+2)^j = \sum_{i=0}^j {j \choose i} \cdot 2^{j-i} \cdot 1^i$$ $$ = (1+2)^j = \sum_{i=0}^j {j \choose i} \cdot 2^{j-i}$$

But I am not sure how to do the combinatorial proof.

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  • $\begingroup$ What do you mean by 'the combinatorial proof'? $\endgroup$ – caverac Dec 21 '18 at 10:36
  • $\begingroup$ @caverac In a combinatorial proof, you count the same set of objects in two different ways to show that the expressions are equal. For instance, to prove the identity $k\binom{n}{k} = n\binom{n - 1}{k - 1}$, you would count committees of size $k$ with a chairperson that can be selected from a group with $n$ people. The left side counts the number of ways of selecting a group of $k$ people, then choosing a chairperson from among the group. The right side counts the number of ways of selecting a chairperson, then selecting the other $k - 1$ members of the committee from the remaining people. $\endgroup$ – N. F. Taussig Dec 21 '18 at 10:46
  • $\begingroup$ @N.F.Taussig Thanks for the explanation, I didn't realize the OP was looking for a proof different to the one he sketched, which is completely valid $\endgroup$ – caverac Dec 21 '18 at 12:20
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A combinatorical proof could go as follows:

  • The number of digit sequences of length $j$ formed with $3$ digits $\{1,2,3\}$ is: $\color{blue}{3^j}$.
  • Now, fix one digit. For example $1$. It can occur $\color{blue}{i=0,..,j}$ times in a digit sequence.
  • The number of ways to place $i$ times the digit $1$ is: $\color{blue}{\binom{j}{i}}$
  • You can fill the remaining $j-i$ places with any of the two other digits: $\color{blue}{2^{j-i}}$ All together: $$\boxed{\color{blue}{\sum_{i=0}^j \binom{j}{i}2^{j-i} = 3^j}}$$
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  • $\begingroup$ I like your use of colour! :) $\endgroup$ – Shaun Dec 21 '18 at 11:27
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The problem - how many $j$-length vectors can be composed of the digits $\{0,1,2\}$?

RHS - straight forward.

LHS - first, pick the $i$-indexes in the vector where 0 appears - $j \choose i$, then choose between $\{1,2\}$ for the $j-i$ remaining indexes - $2^{j-i}$ options of doing so. Summing over all $i$'s gives all the required vectors.

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