The following sum came up in a combinatorial argument. I know what it equals thanks to Wolfram Alpha, but I'm not sure how to show it
$$\sum_{i=0}^n {n\choose i}2^{i}=3^{n}$$
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This question is extremely unclear IMO. What is the $\sum$'s subscript? How could $0^n$ multiplied by anything be equal to anything other than $0$??? – barak manos Jan 15 '16 at 11:15
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@barakmanos it should be clear now? – Stella Biderman Jan 15 '16 at 11:18
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Yes, much better now. I'd use induction BTW... – barak manos Jan 15 '16 at 11:21
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4See http://math.stackexchange.com/questions/525266/prove-sum-binom-nk-2k-3n-using-the-binomial-theorem or http://math.stackexchange.com/questions/258009/how-to-prove-that-sum-r-0n-binomnr2r-3n – Martin Sleziak Feb 01 '16 at 06:33
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Hint:$ $ $ $ $ $
$$3^n = (2+1)^n$$
5xum
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3@StellaBiderman No matter the experience level, time of day and age, things like this will happen all the time. I promise :P – 5xum Jan 15 '16 at 11:22
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For the record, the proof 5xum alludes to goes as follows:
The Binomial Theorem states that $$(a+b)^n=\sum_{I=0}^n{n \choose i} a^ib^{n-i}$$ For $a=2,b=1$, we notice that our sum is of the above form.
Stella Biderman
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As a combinatorial proof, both sides of the equality count the number of partitions of $\{1,2,\ldots,n\}$ into three ordered parts $(A,B,C)$.
Method 1: For each element, we add it to $A$, $B$, or $C$, giving $3^n$ possibilities.
Method 2: We choose $i \in \{0,1,\ldots,n\}$ elements to be added to $B \cup C$, and the rest go into $A$. Of those $i$ elements, we add them to $B$ or $C$ one by one, giving $2^i$ possibilities. This totals $\sum_{i=0}^n \binom{n}{i} 2^i$.
Rebecca J. Stones
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1(Note too that this instantly generalizes to the full binomial theorem, using $m+n$ ordered parts rather than just $1+2$) – Steven Stadnicki Feb 01 '16 at 06:40