A) \[1\]
B) \[i\]
C) \[{{x}^{2}}+{{y}^{2}}\]
D) \[1+{{n}^{2}}\]
Correct Answer: C
Solution :
Given, \[(1+i)\,(2i+1)\,(1+3i)...(1+ni)+x+iy\] ?.(i) \[\because \]\[({{a}_{1}}+i{{b}_{1}})\,({{a}_{2}}+i{{b}_{2}})...({{a}_{n}}+i{{b}_{n}})=A+iB\] \[\Rightarrow \]\[{{r}_{1}}.{{r}_{2}}.....{{r}_{n}}\,[\cos \,({{\theta }_{1}}+{{\theta }_{2}}+.....+{{\theta }_{n}})\] \[+i\,\sin \,({{\theta }_{1}}+{{\theta }_{2}}+.....+{{\theta }_{n}})]\] \[=A+iB\] \[\Rightarrow \] \[{{r}_{1}}.{{r}_{2}}....{{r}_{n}}=\sqrt{{{A}^{2}}+{{B}^{2}}}\] and \[\tan \,({{\theta }_{1}}+{{\theta }_{2}}+.....+{{\theta }_{n}})=\frac{B}{A}\] ?.(iii) Now, from Eq. (ii), we get \[\sqrt{{{1}^{2}}+{{1}^{2}}}.\sqrt{{{2}^{2}}+{{1}^{2}}}.\sqrt{{{3}^{2}}+{{1}^{2}}}.....\sqrt{{{1}^{2}}+{{n}^{2}}}=\sqrt{{{x}^{2}}+{{y}^{2}}}\]\[\Rightarrow \] \[\sqrt{2}.\sqrt{5}.\sqrt{10}....\sqrt{1+{{n}^{2}}}=\sqrt{{{x}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[1.2.5.10....(1+{{n}^{2}})+{{x}^{2}}+{{y}^{2}}\]
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