Question

Find the total area of $12$ squares whose sides are $12cm,13cm,...,23cm$ respectively

Answer 1

To find the total surface area of $12$ squares we have to apply the formula for sum of squares of $n$ natural numbers and that is:

Sum of squares of $n$ natural numbers is $S_n=\frac{n(n+1)(2n+1)}{6}$

The series of total surface area of $12$ squares is as follows:

${12}^2+{13}^2+.......+{23}^2$

We have to find the sum of squares from $1$ up to $23$ and subtract the sum of squares of series from $1$ up to $11$ from that as shown below:

$S=(1^2+2^2+3^2+.......+{23}^2)-(1^2+2^2+3^2+.......+{11}^2)$

$=\frac{23(23+1)\left[\right(2\times 23)+1]}{6}-\frac{11(11+1)\left[\right(2\times 11)+1]}{6}=\frac{(23\times 24)(46+1)}{6}-\frac{(11\times 12)(22+1)}{6}$

$=\frac{(23\times 24\times 47)}{6}-\frac{(11\times 12\times 23)}{6}=(23\times 4\times 47)-(11\times 2\times 23)=4324-506=3818$

Hence, the total surface area is $3818$.