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# 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

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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
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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