Question

If $(1+i)(1+2i)(1+3i)....\left(1_{n}i\right)=a+ib,\text{then}2\times 5\times 10\times ....\times (1+n^2)\text{is equal to}$

• A. $\sqrt{a^2+b^2}$
• B. $\sqrt{a^2-b^2}$
• C. $a^2+b^2$
• D. $a^2-b^2$
• E. a + b

Answer 1

The correct option is C

$a^2+b^2$

$a^2+b^2(1+i)(a+2i)(1+3i)...(1+ni)=a+ib$

Taking modulus on both the sides, we get :

$|(1+i)(1+2i)(1+3i)....(1+ni)|=|a+ib||(1+i)(1+2i)(1+3i)....(1+ni)|\text{can be written as}|(1+i)||1+2i||(1+3i)|...|(1+ni)|\sqrt{1^2+1^2}\times \sqrt{1^2+2^2}\times \sqrt{1^2+3^2}\times .....\times \sqrt{1+n^2}=\sqrt{a^2+b^2}\Rightarrow \sqrt{2}\times \sqrt{5}\times \sqrt{10}\times ....\times \sqrt{1+n^2}=\sqrt{a^2+b^2}$

Squaring on both sides, we get :

$2\times 5\times 10\times ....\times (1+n^2)=a^2+b^2$