Recursion Warmup - Tower Numbers

This is the formation of triangles in the outside layer of a triangle, it means the blocks must increased by $1$ in every layer. To the $5^{th}$ row, the blocks are $\displaystyle 15+(1\times 5)+1 = \boxed{21}$

$15$ is the $5^{th}$ row, and $1\times 5$ is the blocks increased each row times the row, plus $1$ to get the sixth layer.

Those terms of the sequence (except the $0$ ) look like the 3rd pair of diagonals (moving inwards) of the pascal's triangle. Those numbers are called triangular numbers and they can be generated by the formula: $\Large{T_n = \binom{n + 1}{2}}$

Hence for $\large{n = 6}$ we got: $\Large{\binom{n + 1}{2} = \frac{n\cdot (n + 1)}{2} = \frac{6\cdot 7}{2} = \boxed{21}}$

The base of the shape are the key, Let's be n the number of cube in the base, then in the next "floor" would be n-1 cube and so recursively, then the function than represent the number of cube in the shape are $\sum_{i=1}^{n} i$ ; now from n=6 the answer are 21

1+2=3, 1+2+3=6,...,1+2+3+4+5+6=21.

1+2+3+4+5=15; 15+6=21

n 1=1, n (i>1)=i^2-n_(i-1),i ⊂ N