+0

# 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

+1
159
1
+300

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
Sort:

#1
+82557
+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 2n

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

20 + 21 + 22 + 23  = 1 + 2 + 4 + 8  =   15  =  24 - 1

And adding 1 to this sum  = 16  =   24  which is just  the sum of the 5th row elements

CPhill  Sep 26, 2017

### 11 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details