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Trying to determine a formula for the sum of the entries of the $n$th row of Pascal’s triangle, for any natural number $n$. Any proof will do as I have to determine $3$ different proofs.

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So far, I've been working with a proof which includes Pascal's Identity and using combinations to produce $2^n$.

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Expand $(1+1)^n$ by the binomial theorem.

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  • $\begingroup$ Is it sufficient enough to say: $\sum_{k=0}^{n} \binom{n}{k} 1^{n-k}1{^{k}} = (1+1)^{n} = 2^{n}$ ? $\endgroup$ – user115461 Jan 27 '14 at 20:14
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Induction using $$\displaystyle \sum_k x_{n,k} = \sum_k (x_{n-1,k-1} + x_{n-1,k} ) = \sum_k x_{n-1,k-1} + \sum_k x_{n-1,k} = 2 \sum_k x_{n-1,k}$$ starting at $$\displaystyle\sum_k x_{0,k}=1$$.

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