Question

Find the total area of $14$ squares whose sides are $11$ cm, $12$ cm, $13$ cm, ..... , $24$ cm.

Answer 1

Total area of $14$ squares
$={11}^2+{12}^2+{13}^2+....+{24}^2$
$=(1^2+2^2+...+{24}^2)-(1^2+2^2+...+{10}^2)$
$=\sum n^2=\frac{n(n+1)(2n+1)}{6}$
$=\frac{24\times 25\times 49}{6}-\frac{10\times 11\times 21}{6}$
$=4900-385=4515$