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Solve using the binomial theorem:

$$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+\dotsc+2^n\binom{n}{n} $$

I understand the binomial theorem but i am not sure how to apply it to this. Can someone please explain, thank you!

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    $\begingroup$ Think about...$(2+1)^n$ $\endgroup$ – ClassicStyle Nov 7 '14 at 2:27
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The binomial theorem states that $\left(1+x\right)^n = {n\choose 0}x^0 + {n\choose 1}x^1 + \ldots + {n\choose n}x^n$. You're being asked to evaluate an expression that is effectively ${n\choose 0}2^0 + {n\choose 1}2^1 + \ldots + {n\choose n}2^n$. Can you see what the missing step is?

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Notice that

$$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+\dotsc+2^n\binom{n}{n} = \sum\limits_{k=0}^n \binom{n}{k}2^k1^{n-k}.$$

Recall that the binomial theorem states:

$$\sum\limits_{r=0}^n\binom{n}{r}a^rb^{n-r} = (a+b)^n.$$

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