If (1+i)(1+2i)(1+3i)...(1+ni)=a+ib, then 2.5.10.17...(1+n^2)= - Vidyadip

MCQ

If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2.5.10.17...(1+\text{n}^2)=$

A
$\text{a}-\text{ib}$
B
$\text{a}^2-\text{b}^2$
✓
$\text{a}^2+\text{b}^2$
D
none of these

✓

Answer

Correct option: C.

$\text{a}^2+\text{b}^2$

$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get,
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$

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