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Find the sum of the binomial coefficients $(^n_0)+(^n_1)+...+(^n_{n-1})+ (^n_n).$

I'm not good with the binomial theorem but I do know that $\sum_{k=0}^{n} (^n_k)x^{k} = (^n_0)+(^n_1)x+...+ (^n_n)x^n.$

Now comparing the binomial coefficients I can see some similarities, but the difference would be that $(^n_{n-1})$ and the $x^n$.

If I just have $\sum_{k=0}^{n} (^n_k) = (^n_0)+(^n_1)+...+ (^n_n).$ then all I would be needing would be that $(^n_{n-1})$.

Any help?

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    Write the expansion $(1+x)^n$. – 2017-02-03
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    $\sum_{k=0}^{n} (^n_k)x^{k}$ – 2017-02-03
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    Set $x=1$ in it. – 2017-02-03
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    The you would get something like this? $2^n$? – 2017-02-03
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    OK. and the first side is $(1+1)^n$.no? – 2017-02-03
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    You can also try a few examples with a small $n$ and then induce the general formula. Also, that sum is equal to the amount of all subsets with cardinality from 0 to n, of a set with n elements... which is the cardinality of the powerset, if you have some knowledge of elementary set theory. – 2017-02-03
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    Come on, $\binom n{n-1}$ is in the $\dots$ ! – 2017-02-03
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    Yes we would get $(1+1)^n$ which will equal to $2^n$ – 2017-02-03

2 Answers 2

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Do you know that $(1+x)^n = \sum_{k=0}^{n} (^n_k)1^{n-k}x^{k} = (^n_0)+(^n_1)x+...+ (^n_n)x^n $? We need $x=1$ here.

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    Yes, but I didn't add the $(1+x)^n$, so it would be then $(2^n)$ right? – 2017-02-03
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    @FernandoFernandez sure, take $x=1$ and we get $2^n$ as the answer to the sum. – 2017-02-03
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    Just out of curiousity how would $2^n$ be denoted in summation form? I struggle with binomial theorem, so i want to understand it a little bit better. – 2017-02-03
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    Why would you want that? We just wrote it as a sum of binomial coefficients. The sum is now simplified to one term. – 2017-02-03
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    I see, do you have any book or website to recommend where I can get more familiar with summation proofs and the binomial theorem? – 2017-02-03
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If $X$ is a set with $\# X =n$, then $\binom{n}{0}$ is the number of subsets with no elements, $\binom{n}{1}$ is the number of subsets with one element and so on. So $\sum_{j=0}^n \binom{n}{j}$ gives the number of all subsets of $X$ which is $2^n$ since $\# \mathcal P (X)=2^n$.

[This is why some people write $2^X$ for the powerset of $X$ instead $\mathcal P(X)$.]