Astfel vedem că numărul total de pătrate pe o grilă $N × M$ este:

<tex> \[\sum_{L=2}^{N} {(N-L+1) \cdot (N-L+1) \cdot (L-1)}\] = \[\sum_{L=1}^{N-1} {(N-L) \cdot (N-L) \cdot L}\] = \[\sum_{L=1}^{N-1} {L \cdot N^2 - 2N \cdot L^2 + L^3}\] = N^2 \cdot \[\sum_{L=1}^{N-1} {L}\] - 2N \cdot \[\sum_{L=1}^N-1 {L^2}\] + \[\sum_{L=1}^{N-1} {L^3}\] = N^2 \cdot \frac {N(N-1)} {2} - 2N \cdot \frac {N(N+1)(2N+1)} {6} + {\frac {N(N+1)} {2}}^2 </tex>
 

Sumă după $L$ de la $2$ la $N$ de $(N–L+1) × (N–L+1) × (L–1) =$
suma $L$ de la $1$ la $N-1$ de $(N–L) × (N–L) × L =$ suma cu $L$ de la $1$ la $N–1$ de $LN^2^–2 NL^2^ + L^3^ = N^2^ ×$ suma $L$ de la $1$ la $N-1-2N$ suma $L$ de la $1$ la $N–1$ de $L^2^ +$ suma de $L$ de la $1$ la $N–1$ de $L^3^ = N^2^ ×  N(N-1)/2-2N n(n-1)(2n-1)/6 + [n(n-1)/2]^2^$.

<tex>
\[ \sum_{L=2}^N {(N-L+1)(N-L+1)(L-1)} \] =
</tex>
<tex>
= \[\sum_{L=1}^{N-1} {(N-1)(N-1)L} \] =
</tex>
<tex>
 = \[\sum_{L=1}^{N-1} {LN^2 - 2NL^2 + L^3}\] =
</tex>
<tex>
= \[N^2\sum_{L=1}^{N-1} {L} - 2N\sum_{L=1}^{N-1} {L^2} + \sum_{L=1}^{N-1} {L^3} \] =
</tex>
<tex>
 = \[n^2\fraq {N(N-1)}{2} - 2n\fraq{N(N-1)(2N-1)}{6} + \left[\fraq{N(N-1)}{2}\rigt]^2 .\]
</tex>
 

Am folosit formulele cunoscute din manualul de clasa a 10-a:
 <tex> \sum_{k=1}^n k = \frac {n(n+1)} {2} </tex>
 <tex> \sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)} {6} </tex>

 <tex> \sum_{k=1}^n k^3 = {\frac {n(n+1)} {2}}^2 </tex>

 <tex> \sum_{k=1}^n k^3 = \left[\frac {n(n+1)} {2} \right]^2 </tex>

Această soluţie are complexitatea $O(1)$, fiind o simplă formulă.