Question

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

Answer 1

The areas of the squares form the series ${11}^2+{12}^2+...+{24}^2$
Total area of $14$ squares $={11}^2+{12}^2+{13}^2+....+{24}^2$
$=(1^2+2^2+3^2+....+{24}^2)-(1^2+2^2+3^2+...+{10}^2)$
$=\sum _1^{24}{n}^2-\sum _1^{10}{n}^2$
$=\frac{24(24+1)(48+1)}{6}-\frac{10(10+1)(20+1)}{6}$
$=\frac{\left(24\right)\left(25\right)\left(49\right)}{6}-\frac{\left(10\right)\left(11\right)\left(21\right)}{6}$
$=4900-385$
$=4515$ sq. cm