Question

If $(1+i)(1+2i)(1+3i)...(1+ni)=a+ib,\mathrm{then}2\times 5\times 10\times ....\times (1+n^2)$ is equal to
(a) $\sqrt{a^2+b^2}$
(b) $\sqrt{a^2-b^2}$
(c) $a^2+b^2$
(d) $a^2-b^2$
(e) $a+b$

Answer 1

(c) a² + b²

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get:
$$\begin{gathered} \left|\left(1+i\right)\left(1+2i\right)\left(1+3i\right).......\left(1+ni\right)\right|=\left|a+ib\right| \\ \left|\left(1+i\right)\left(1+2i\right)\left(1+3i\right).......\left(1+ni\right)\right|\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\left|\left(1+i\right)\right|\left|\left(1+2i\right)\right|\left|\left(1+3i\right)\right|.......\left|\left(1+ni\right)\right| \end{gathered}$$

$\sqrt{1^2+1^2}\times \sqrt{1^2+2^2}\times \sqrt{1^2+3^2}\times ...\times \sqrt{1+n^2}=\sqrt{a^2+b^2}$

$\Rightarrow \sqrt{2}\times \sqrt{5}\times \sqrt{10}\times ...\times \sqrt{1+n^2}=\sqrt{a^2+b^2}$

Squaring on both the sides, we get:

$2\times 5\times 10\times ...\times (1+n^2)=a^2+b^2$
2