Prove that the sum in each row of a Pascal triangle is double that of the previous row I'm trying to prove that the sum of every row in Pascal triangle is double the previous row by using Pascal's rule:  
$${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}.$$
It's easy for me to understand why is it correct.  We can use the rule in order to replace any element to 2 elements from the previous row and to show that way that each row has exactly the same elements of the previous row times $x$.  However, I'm struggling to write the general proof.
 A: It's  not very long with Pascal's relation. You have to know that the first and last coefficients in any row are equal to $1$:
\begin{align}
\sum_{k=0}^{n}\binom nk&=1+\sum_{k=1}^{n-1}\binom nk +1 \\
&=1+\sum_{k=1}^{n-1}\binom{n-1}k+\sum_{k=1}^{n-1}\binom{n-1}{k-1}+1 \\
& = 1+\sum_{k=1}^{n-1}\binom{n-1}k+\sum_{k=0}^{n-2}\binom{n-1}{k}+1 \\
&=\sum_{k=0}^{n-1}\binom{n-1}k+\sum_{k=0}^{n-1}\binom{n-1}{k}=2\sum_{k=0}^{n-1}\binom{n-1}{k} %Making the whole thing readable.
\end{align}
A: by the definition of the Pascal triangle, every number is the sum of the two numbers above it. also, every number is above two numbers in the row below it. therefore, every number summed twice in the next row, which cause the sum of a row to be double the sum of the previous one.
A: Try proving that the elements of row $k$ sum to $2^k$. (The row with a 1 is row 0)
Further Hint: Binomial Theorem.
A: Not a complete solution by any means, but for clarity and to assist a full solution:
You are trying to show
$$\sum_{k=0}^n\binom{n}{k}=\frac12\sum_{k=0}^{n+1}\binom{n+1}{k}.$$
Perhaps consider
$$\binom{n}{k}=\frac{n!}{k!\ (n-k)!}$$
and
$$\binom{n+1}{k}=\frac{(n+1)!}{k!\ (n-k+1)!}=\frac{(n+1)\cdot n!}{k!\ (n-k+1) (n-k)!}$$
