If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib,


If $(1+i)(1+2 i)(1+3 i) \ldots(1+n i)=a+i b$, then $2.5 .10 .17 \ldots \ldots . .\left(1+n^{2}\right)=$

(a) a − ib

(b) a2 − b2

(c) a2 + b2

(d) none of these


(c) $a^{2}+b^{2}$

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

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 wriiten as $|(1+i)||(1+2 i)||(1+3 i)| \ldots|(1+n i)|$

$\therefore \sqrt{1^{2}+1^{2}} \times \sqrt{1^{2}+2^{2}} \times \sqrt{1^{2}+3^{2}} \ldots \times \sqrt{1^{2}+n^{2}}=\sqrt{a^{2}+b^{2}}$

$\Rightarrow \sqrt{2} \times \sqrt{5} \times \sqrt{10} \ldots \times \sqrt{1+n^{2}}=\sqrt{a^{2}+b^{2}}$

Squaring on both the sides, we get:

$2 \times 5 \times 10 \ldots \times\left(1+n^{2}\right)=a^{2}+b^{2}$

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