Question

If $ (1+i)\,(2i+1)\,(1+3i)....(1+ni)=x+iy, $ then $ 2.5.10....(1+{{n}^{2}}) $ is equal to

• A. $ 1 $
• B. $ i $
• C. $ {{x}^{2}}+{{y}^{2}} $
• D. $ 1+{{n}^{2}} $

Answer 1

Answer: (C)

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}} $