Simply Awsome!

Algebra Level 4

If,

$\sum _{ i=1 }^{ 200 }{ \sqrt { \left( 1+\frac { 1 }{ { i }^{ 2 } } +\frac { 1 }{ { \left( 1+i \right) }^{ 2 } } \right) } } =\frac { A }{ B }$

Find: $A+B$ where $A$ and $B$ are coprime positive integers.

Dedicated to all my 200+ followers

Thanks a lot for encouraging me to improve!

The answer is 40601.

1 solution

Aditya Kumar
Aug 31, 2015

$1+\frac { 1 }{ { i }^{ 2 } } +\frac { 1 }{ { \left( 1+i \right) }^{ 2 } } =1+\frac { 1 }{ { i }^{ 2 } } +\frac { 1 }{ { \left( 1+i \right) }^{ 2 } } +\frac { 2 }{ i\left( i+1 \right) } -\frac { 2 }{ i\left( i+1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =1+2\left( \frac { 1 }{ i } -\frac { 1 }{ i+1 } \right) +{ \left( \frac { 1 }{ i } -\frac { 1 }{ i+1 } \right) }^{ 2 }\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ \left( 1+\frac { 1 }{ i } -\frac { 1 }{ i+1 } \right) }^{ 2 }\\ \therefore \sum _{ i=1 }^{ 200 }{ \sqrt { \left( 1+\frac { 1 }{ { i }^{ 2 } } +\frac { 1 }{ { \left( 1+i \right) }^{ 2 } } \right) } } =\sum _{ i=1 }^{ 200 }{ \left( 1+\frac { 1 }{ i } -\frac { 1 }{ i+1 } \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =201-\frac { 1 }{ 201 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 40400 }{ 201 }$

I did it in exactly same way.

Surya Prakash - 5 years, 9 months ago

Nice. It is a very easy problem. I wanted more people to solve my problems. That's y i posted it.

Aditya Kumar - 5 years, 9 months ago

Something similar I had posted - A Complicated Logarithmic Limit! :P

Satyajit Mohanty - 5 years, 9 months ago

Simply Awsome!

The answer is 40601.

1 solution

0 pending reports