Count 'em All 5!

${50} \times {39}+{49} \times {38}+{48} \times {37}+ \dots + {13} \times {2}+ {12} \times {1}$ Since the answer is a number equal to the above expression, here is a C++ code script that may help a bit:

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    int S=0;
    for(int i=50, j=39; i>0 && j>0; i--,j--){
        S+=(i*j);
    }
    cout<<S<<endl;

From this note , we know that the number of squares in an $a\times b$ grid given that $a\geqslant b$ is $\frac{b(b+1)(3a-b+1)}{6}$

In this case, $a=50$ and $b=39$ , substitute it into the above equation we could get $29120$ .

No. of squares = $\displaystyle \sum_{n=12}^{50}$ = $29120$ . $_\square$