Question

There are 8436 steel balls, each with a radius of 1cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth layer and so on. The number of horizontal layers in the pile is ___.

Answer 1

Let $S_n$ denote the number of balls in row n.
Therefore, $S_2-S_1=2$
$S_3-S_2=3S_4-S_3=4.........S_n=S_{n-1}=n$
Adding all the equations we get = $S_n-S_1=2+3+4+....+n.$
Since $S_1=1,S_n=1+2+3+...+n=\frac{n(n+1)}{2}.$
Therefore total number of balls -sum of balls in all the row
$=\frac{\sum n(n+1)}{2}=\frac{1}{2}(\sum n^2+\sum n)=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}$
Given that total number of balls -$8436\to \frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}=8436\to n=36$.
Hence, there are 36 layers in the pile.