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I have to prove that $$\sum_{i=0}^n {{n \choose i} \times 2^i} = 3^n $$ Such that ${n \choose i} = \frac{n!}{i!(n-i)!} $ and $n $ is some arbitrary int I proved we can expand 2^I in a way such that $$2^i = {i \choose 0} +{i \choose 1} + ... + {i \choose i} $$ But this doesn't seem to approach the proof is want so I tried to expand out ${n \choose i}$ and again this got ugly and led no where. I can't quite seem to find any approach that would lead to $3$ of any power.

I know this proof is true for the base case$$\sum_{i=0}^0 {{0 \choose i} \times 2^i} = 1 => {0 \choose 0} \times 2^0 = 1 => 1 = 1 $$ Could anyone give me a hint to lead me in the right direction?

EDIT: Wow, I always feel so stupid when I miss tiny identities such as that! Thank you everyone! I got my answer, I wish I could give all of you checks for the advice.

spstephens
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3 Answers3

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In the binomial theorem, $$(x+y)^n={n \choose 0}x^{n-0}y^0+{n \choose 1}x^{n-1}y^1+{n \choose 2}x^{n-2}y^2+\cdots+{n \choose n}x^{n-n}y^n=\sum_{i=1}^{n} {n \choose i} x^{n-i}y^i$$

Put $x=1$ and $y=2$.

SchrodingersCat
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we have $$\left(1 + x\right)^n = 1 + {n \choose 1} x + {n \choose 2} x^2+ \cdots + x^n \tag 1$$ now putting $x = 2$ in $(1)$ gives you the result.

abel
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Consider a combinatorial argument. Let $\Lambda = \{a, b, c\}$ be our alphabet. We select the $i$ positions in an $n$ letter word for $\{a, b\}$ characters in $\binom{n}{i}$ ways and multiply by $2^{i}$ as we have a word (is each selected slot a $a$ or is it a $b$?). The remaining $n-i$ slots are all $c$'s. Now we add up over all possible values of $i$:

$$\sum_{i=0}^{n} \binom{n}{i} 2^{i}$$

Observe that we have enumerated all strings in $\Lambda^{n}$. There are $3^{n}$ such strings.

ml0105
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