The sum of the areas of the 10 squares,the le | vedclass

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(A) The sum of the areas of the squares is given by the sum of the squares of their side lengths:Sum $= 20^2 + 21^2 + \ldots + 29^2$Using the formula

Vedclass · Accepted Answer

$6085$

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(A) The sum of the areas of the squares is given by the sum of the squares of their side lengths:Sum $= 20^2 + 21^2 + \ldots + 29^2$Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we can write:Sum $= \sum_{k=1}^{29} k^2 - \sum_{k=1}^{19} k^2$$= \frac{29(29+1)(2 	imes 29 + 1)}{6} - \frac{19(19+1)(2 	imes 19 + 1)}{6}$$= \frac{29 	imes 30 	imes 59}{6} - \frac{19 	imes 20 	imes 39}{6}$$= (29 	imes 5 	imes 59) - (19 	imes 10 	imes 13)$$= 8555 - 2470 = 6085 \, cm^2$.

The sum of the areas of the $10$ squares,the lengths of whose sides are $20 \, cm, 21 \, cm, \ldots, 29 \, cm$ respectively is ...... $cm^2$.

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