Squares n Circles

Let the side of first square be ${ a }_{ 1 }$ .

$\sqrt { 2 } { a }_{ 1 }=2R$

${ a }_{ 1 }=\sqrt { 2 } R$

Let ${ a }_{ 2 }$ be the side of second square.

$\sqrt { 2 } { a }_{ 2 }={ a }_{ 1 }$ ( $\because$ Radius of second circle=a/2)

${ a }_{ 2 }=\cfrac { { a }_{ 1 } }{ \sqrt { 2 } }$

Similarly,

${ a }_{ 3 }=\cfrac { { a }_{ 2 } }{ \sqrt { 2 } } =\cfrac { { a }_{ 1 } }{ \sqrt { 2 } }$

..............................................

.............................................

Sum of areas of squares as $n\rightarrow \infty$ ,

${ S }_{ \infty }={ a }_{ 1 }^{ 2 }\quad +\quad { a }_{ 2 }^{ 2 }\quad +\quad { a }_{ 3 }^{ 2 }\quad +\quad .\quad .\quad .\quad \infty$

${ S }_{ \infty }\quad ={ a }_{ 1 }^{ 2 }\quad +\quad \cfrac { { a }_{ 1 }^{ 2 } }{ 2 } \quad +\quad \cfrac { { a }_{ 1 }^{ 2 } }{ 4 } \quad +\quad .\quad .\quad .\quad \infty$

${ S }_{ \infty }\quad ={ a }_{ 1 }^{ 2 }(1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\quad .\quad .\quad .\quad \infty )$

This is an infinite GP with common difference 1/2.

${ S }_{ \infty }\quad ={ a }_{ 1 }^{ 2 }(\cfrac { 1 }{ 1-\frac { 1 }{ 2 } } )$

${ S }_{ \infty }\quad =2{ a }_{ 1 }^{ 2 }\quad =4{ R }^{ 2 }$

Hence the answer is $4{ R }^{ 2 }$ .