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# If $\;(2+i)(2+2i)(2+3i)....(2+ni)=x+iy\;$ , then$\;5\;.8\;.13....(4+n^{2})=$----

$(a)\;x^{2}+y^{2} \qquad(b)\;x^{2}-y^{2} \qquad(c)\;2(x^{2}+y^{2}) \qquad(d)\;0$

Answer : $\;x^{2}+y^{2}$
Explanation :
Given that , $\;(2+i)(2+2i)(2+3i)....(2+ni)=x+iy\;$---(1)
$\overline{(2+i)}\;\overline{(2+2i)}\;\overline{(2+3i)}....\overline{(2+ni)}=\overline{x+iy}\;=x-iy$
$\;(2-i)(2-2i)(2-3i)....(2-ni)=x-iy\;$----(2)
Multiplying (1) and (2) , we get $\;5\;.8\;.13....(4+n^{2})=x^{2} +y^{2}$