Question

A square the is having side of length 15 cm how many such tiles would be required to cover the square floor of a bathroom of sides 3m?

Answer 1

$\frac{3}{6}+\frac{3.5}{6.9}+\frac{\mathrm{3.5.7}}{\mathrm{6.9.12}}+........\infty$

$=3+1+\frac{1.3}{6}+\frac{\mathrm{1.3.5}}{6.9}+\frac{\mathrm{1.3.5.7}}{\mathrm{6.9.12}}+.....\infty -4$

$=3[1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+....\infty ]-4$

$=3[1+\left(\frac{-1}{2}\right)(-\frac{2}{3})+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}{(-\frac{2}{3})}^2+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}{(-\frac{2}{3})}^3+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2)(-\frac{1}{2}-3)}{4!}{(-\frac{2}{3})}^4+....\infty ]-4$

$=3{[1-\frac{2}{3}]}^{-\frac{1}{2}}-4$

$=3\sqrt{3}-4$