Question

If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n²)=
(a) a − ib
(b) a² − b²
(c) a² + b²
(d) none of these

Answer 1

(c) a² + b²

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

Taking modulus on both the sides, we get,

$$\begin{gathered} \left|\left(1+i\right)\left(1+2i\right)\left(1+3i\right)......\left(1+ni\right)\right|=a+ib \\ \left|\left(1+i\right)\left(1+2i\right)\left(1+3i\right)......\left(1+ni\right)\right|\mathrm{can}\mathrm{be}\mathrm{wriiten}\mathrm{as}\left|\left(1+i\right)\right|\left|\left(1+2i\right)\right|\left|\left(1+3i\right)\right|....\left|\left(1+ni\right)\right| \\ \therefore \sqrt{1^2+1^2}\times \sqrt{1^2+2^2}\times \sqrt{1^2+3^2}....\times \sqrt{1^2+n^2}=\sqrt{a^2+b^2} \end{gathered}$$

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

Squaring on both the sides, we get:

$2\times 5\times 10....\times (1+n^2)=a^2+b^2$
2×5×10×.....(1 + n2) = a2 + b2