What is the sum of the squares of the digits from 1 to 9? · ABC26GN0071

We know that: $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)$. $\therefore\left(1^{2}+2^{2}+3^{2}+\ldots \ldots \ldots+9^{2}\right)=\left(\frac{1}{6} \times 9 \times 10 \times 19\right)=285$.