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.
Best Answer
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}