Question

If $\alpha =1/4$ and $P_n$ denotes the perimeter of the nth square then${\sum }_{n=1}^{\infty }{P}_n$ equals

• A. $\frac{8}{3}(4+\sqrt{10})$
• B. $\frac{8}{3}$
• C. $\frac{16}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{8}{3}(4+\sqrt{10})$

Parameter of $n^{th}$ square $=4a_n=4a_1{r}^{n-1}$

${\sum }_{n=1}^{\infty }$$=P_1+P_2+P_3+....+P_n=4a_1(1+r+r^2+r^3+....infty)$

Sum of infinite GP $=\frac{a}{1-r}$
${\sum }_{n=1}^{\infty }=4a_1\frac{1}{1-r}=\frac{4a_1}{1-\sqrt{2{\alpha }^2-2/alpha+1}}$

Given $a_1=1,\alpha =\frac{1}{4}$

${\sum }_{n=1}^{\infty }$$=\frac{4}{1-\sqrt{\frac{2}{16}-\frac{2}{4}+1}}=\frac{4}{1-\frac{\sqrt{2-8+16}}{4}}=\frac{16}{4-\sqrt{10}}=\frac{16}{4-\sqrt{10}}\times \frac{4+\sqrt{10}}{4+\sqrt{10}}=\frac{16(4+\sqrt{10})}{6}=\frac{8}{3}(4+\sqrt{10})$