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If $(1+i)(1+2i)(1+3i).....(1+ni)=x+iy$, then $2.5.10.....(1+n^2)$ = ?

$\begin{array}{1 1}(A) \; x^2+y^2 \\(B)\;x^2-y^2 \\(C)\;(x+iy)^2 \\(D)\; (x-iy)^2 \end{array}$

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Toolbox:
  • $|x+iy|=\sqrt{x^2+y^2}$
Given $(1+i)(1+2i)(1+3i).............(1+ni)=x+iy$
$\Rightarrow\:|(1+i)(1+2i)(1+3i).............(1+ni)|=|x+iy|$
$\Rightarrow\:|1+i||1+2i|..........|1+ni|=\sqrt {x^2+y^2}$
$\Rightarrow\:\sqrt 2\sqrt 5\sqrt {10}.......\sqrt{1+n^2}=\sqrt {x^2+y^2}$
squaring on both the sides we get
$2.5.10........(1+n^2)=x^2+y^2$
answered Jul 25, 2013 by rvidyagovindarajan_1
 

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