N cube build x layers

Assume the sequence $\{a_n\}_{n=1}^{\infty}$ to be the number of cubes in each level. then, the following recurrence relation is true.

$a_n=a_{n-1}+n+1$

the corresponding generating function is

$f(x)=\frac{1}{(1-x)^3}$

the generating function that has $b_m=\sum_{i=0}^{m}a_i$ as its coefficient is (we are adding the number of cubes for $m$ levels)

$g(x)=\frac{f(x)}{1-x}=\frac{1}{(1-x)^4}$

if the Taylor expansion of $g(x)$ , around $0$ , is written, one finds out that the coefficients of $g(x)$ are $b_m=\frac{(3+m)!}{m!3!}= \binom{m+3} {3}$ . Then we need to find the maximum $m$ for which

$b_m= \binom{m+3}{3} \leq 1000$

it turns out to be $m=16$ , which means that we can have 17 levels. Also, $\binom{16+3}{3} = \binom{19}{3}=969$ , so the remaining of the cubes can be calculated.

$1000- \displaystyle \sum _{ j=1 }^{ 17 }{\displaystyle \sum _{ k=1 }^{ j }{ k } } =31$

..so, answer is $17.31$