If $(1+i)(1+2 i)(1+3 i) \ldots(1+n i)=a+i b$, then $2 \times 5 \times 10 \times \ldots \times\left(1+n^{2}\right)$ 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) $a+b$
(c) $a^{2}+b^{2}$
$(1+i)(1+2 i)(1+3 i) \ldots \ldots(1+n i)=a+i b$
Taking modulus on both the sides, we get:
$|(1+i)(1+2 i)(1+3 i) \ldots \ldots(1+n i)|=|a+i b|$
$|(1+i)(1+2 i)(1+3 i) \ldots \ldots(1+n i)|$ can be written as $|(1+i)||(1+2 i)||(1+3 i)| \ldots \ldots \ldots|(1+n i)|$
$\sqrt{1^{2}+1^{2}} \times \sqrt{1^{2}+2^{2}} \times \sqrt{1^{2}+3^{2}} \times \ldots \times \sqrt{1+n^{2}}=\sqrt{a^{2}+b^{2}}$
$\Rightarrow \sqrt{2} \times \sqrt{5} \times \sqrt{10} \times \ldots \times \sqrt{1+n^{2}}=\sqrt{a^{2}+b^{2}}$
Squaring on both the sides, we get:
$2 \times 5 \times 10 \times \ldots \times\left(1+n^{2}\right)=a^{2}+b^{2}$