\[\begin{array}{l}
Ta\,\,co\,\,cong\,\,thuc:\\
{1^2} + {2^2} + {3^2} + ... + {n^2} + {(n + 1)^2} = \frac{{\left( {n + 1} \right)\left( {n + 2} \right)(2n + 3)}}{6}\\
{1^2} + {2^2} + {3^2} + ... + {(n - 11)^2} + {(n - 10)^2} = \frac{{\left( {n - 10} \right)\left( {n - 9} \right)(2n - 19)}}{6}\\
Dat\,\,A\,la\,\,tong\,\,can\,\,tim,\,\,ta\,\,co:\\
A = \frac{{\left( {n + 1} \right)\left( {n + 2} \right)(2n + 3)}}{6} - \frac{{\left( {n - 10} \right)\left( {n - 9} \right)(2n - 19)}}{6}\\
= \frac{{\left( {n + 1} \right)\left( {n + 2} \right)(2n + 3) - \left( {n - 10} \right)\left( {n - 9} \right)(2n - 19)}}{6}\\
= \frac{{2{n^3} + 9{n^2} + 13n + 6 - 2{n^3} + 57{n^2} - 541n + 1710}}{6}\\
= \frac{{66{n^2} - 528n + 1716}}{6} = 11{n^2} - 88n + 286 = 11({n^2} - 8n + 26)\\
= > 11({n^2} - 8n + 26) = 121{k^2} < 10000 = > k \le 9\\
ma\,\,n \ge 12 = > k \ge 3\\
3 \le k \le 9\\
Bam\,\,Casio....\\
= > n = 27 = > A = {18^2} + {19^2} + {20^2} + {21^2} + {22^2} + {23^2} + {24^2} + {25^2} + {26^2} + {27^2} + {28^2} = 5929 = {77^2}
\end{array}\]
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