Since the [tex]n[/tex]-th level requires [tex]n[/tex] block, she'll need a total of
[tex]1+2+3+4+5+6+7+8+9+10+11+12[/tex]
blocks. This computation can be done quickly if one remembers that the sum of the first [tex]n[/tex] numbers equals
[tex]\dfrac{n(n+1)}{2}[/tex]
Plugging [tex]n=12[/tex] we get
[tex]\dfrac{12\cdot 13}{2}=6\cdot 13 = 78[/tex]
