We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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**+3 **

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