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 $(1-i^2)(1-4i^2)(1-9i^2)....(1-n^2{i}^2)$ $=a^2+b^2$ $\Rightarrow$ $(1+1)(1+4)(1+9)....(1+n^2)=a^2+b^2$ $\Rightarrow$ $\mathrm{2.5.10......}(1+n^2)=a^2+b^2$