Question

If $ (1+i)(1+2i)(1+3i)....(1+ni)=a+ib, $ then $ 2\times 5\times 10\times .....\times (1+{{n}^{2}}) $ is equal to:

• A. $ {{a}^{2}}+{{b}^{2}} $
• B. $ \sqrt{{{a}^{2}}+{{b}^{2}}} $
• C. $ \sqrt{{{a}^{2}}-{{b}^{2}}} $
• D. $ {{a}^{2}}-{{b}^{2}} $
• E. $ a+b $

Answer 1

Answer: (A)

$ \because $ $ (1+i)(1+2i)(1+3i)....(1+{{n}_{i}})=a+ib $ ...(i) $ \Rightarrow $ $ (1-i)(1-2i)(1-3i)....(1-ni)=a-ib $ ...(ii) $ \therefore $ From Eqs. (i) and (ii), we get $ (1-{{i}^{2}})(1-4{{i}^{2}})(1-9{{i}^{2}})....(1-{{n}^{2}}{{i}^{2}}) $ $ ={{a}^{2}}+{{b}^{2}} $ $ \Rightarrow $ $ (1+1)(1+4)(1+9)....(1+{{n}^{2}})={{a}^{2}}+{{b}^{2}} $ $ \Rightarrow $ $ 2.5.10......(1+{{n}^{2}})={{a}^{2}}+{{b}^{2}} $