NIMCET 2018 Question Paper with solutions

Number of balls in 1st layer = 1
Number of balls in 2nd layer =3=1+2
Number of balls in 3rd layer = 6=1+2 + 3
Number of balls in 4th layer =10=1+2+3+4
Now, the nth term = $\frac{\mathit{n}\left(\mathit{n}+1\right)}{2}$ = $\frac{{\mathit{n}}^2}{2}+\frac{\mathit{n}}{2}$
Sum of n terms of series = $\frac{1}{2}\left[\mathit{\Sigma }{\mathit{n}}^2+\mathit{\Sigma }\mathit{n}\right]$
= $\frac{1}{2}\left[\frac{\mathit{n}\left(\mathit{n}+1\right)\left(2\mathit{n}+1\right)}{6}+\frac{\mathit{n}\left(\mathit{n}+1\right)}{2}\right]$
= $\frac{\mathit{n}\left(\mathit{n}+1\right)\left(\mathit{n}+2\right)}{6}$
Now ,
$\frac{\mathit{n}\left(\mathit{n}+1\right)\left(\mathit{n}+2\right)}{6}$ = 8436
or n (n + 1) (n + 2) = 50616
The above condition is satisfied only for n = 36