For each integer $n$, let $f(n)$ be the sum of the elements of the $n$th row (i.e. the row with $n+1$ elements) of Pascal's triangle minus the sum of all the elements from previous rows. For example, \([f(2) = \underbrace{(1 + 2 + 1)}_{\text{2nd row}} - \underbrace{(1 + 1 + 1)}_{\text{0th and 1st rows}} = 1\). \]What is the minimum value of $f(n)$ for$\(n \ge 2015\)?

RektTheNoob
Sep 26, 2017

#1**+2 **

The sum of the elements of any row [ after the first one] is always 1 greater than the sum of the elements of all the previous rows

To see this....note that the sum of elements on any row [ starting with n = 0] is just 2^{n}

Note that, for example, the sum of the first 4 rows is

2^{0} + 2^{1} + 2^{2} + 2^{3} = 1 + 2 + 4 + 8 = 15 = 2^{4} - 1

And adding 1 to this sum = 16 = 2^{4} which is just the sum of the 5th row elements

CPhill
Sep 26, 2017