0
$\begingroup$

Consider a random number pyramide (see e.g. http://www.puzzle-magazine.com/numbertower.jpg).

Assume that the numbers in the bottom line are equidistantly distributed. Let $s$ denote the left-most number in the bottom line and let $d$ denote the distance between the numbers in the bottom line.

By experiment I have then found that the number $N$ in the $m^{\text{th}}$ box (counted from left) in the $n^{\text{th}}$ row (counted from the bottom) is given by

\begin{equation} N(m,n) = (2s + d(2m + n - 3)) \cdot 2^{n - 2}. \end{equation}

How do I prove that this is correct (if it indeed is correct)?

$\endgroup$
  • 1
    $\begingroup$ Maybe by induction. The base case would be that the bottom row $n=1$ satisfies indeed the formula. Then you would need to show that the $n+1$ row also satisfies the formula assuming it is true for $n$. and using the fact that, by definition of a "pyramid" we must have $N(m,n+1) = N(m,n) + N(m+1,n)$. More precisely : Using this method $\endgroup$ – Zubzub Feb 16 '17 at 13:43
  • $\begingroup$ @Zubzub That was the way to do it. Thank you for the hint. $\endgroup$ – Svend Tveskæg Feb 16 '17 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.