Sum of a row of Pascal's triangle

Text

Prove that the sum of all the binomial coefficients in row $n$ of Pascal’s triangle equals $2^{n}$, i.e. $\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n} = 2^{n},$ and compute this sum for $n=10$.

Solution

We start from the binomial theorem expansion: $(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{k}b^{\,n-k}.$ Setting $a=b=1$ the left side becomes $2^{n}$, while the right side becomes exactly the sum of the coefficients in the row: $\sum_{k=0}^{n}\binom{n}{k} = 2^{n}.$ Combinatorial interpretation: $2^{n}$ is the total number of subsets of an $n$-element set, and $\binom{n}{k}$ counts those with exactly $k$ elements; summing over all $k$ yields all subsets. For $n=10$: $S_{10} = 2^{10} = \boxed{1024}.$