• # question_answer If $(1+i)(1+2i)(1+3i).....(1+ni)=a+ib$, then        2.5.10....$(1+{{n}^{2}})$  is equal to  [Karnataka CET 2002; Kerala (Engg.) 2002] A) ${{a}^{2}}-{{b}^{2}}$ B) ${{a}^{2}}+{{b}^{2}}$ C) $\sqrt{{{a}^{2}}+{{b}^{2}}}$ D) $\sqrt{{{a}^{2}}-{{b}^{2}}}$

We have $(1+i)(1+2i)(1+3i).....(1+ni)=a+ib$           .....(i) Þ $(1-i)(1-2i)(1-3i).....(1-ni)=a-ib$       .....(ii) Multiplying (i) and (ii), we get $2.5.....(1+{{n}^{2}})={{a}^{2}}+{{b}^{2}}$