Question

If $ (1 + i) (1 + 2i) (1 + 3i) ..... (1 + ni) = x + iy,$ then $2\cdot 5 \cdot 10$ $(1 + n^2) =.......$

• A. $1$
• B. $i$
• C. $1+n^2$
• D. $x^2 + y^2$

Answer 1

Answer: (D)

Since $(1 + i) (1 + 2i) (1 + 3i)......(1 + ni)$
$=x+iy \quad \ldots(1)$
$\therefore (1 - i) (1 - 2i) (1 - 3i) .... (1 - ni)$
$=x-iy\quad \ldots(2)$
Multilying $(1)$ and $(2)$ we get,
$\left(1-i^{2}\right)\left(1-4i^{2}\right)\left(1-9i^{2}\right).....\left(1-n^{2}i^{2}\right)$
$=x^{2}-i^{2}y^{2}$
$\Rightarrow \left(1+1\right)\left(1+4\right)\left(1+9\right)....\left(1+n^{2}\right)=x^{2}+y^{2}$
$\Rightarrow 2\cdot5\cdot10 .....+\left(1+n^{2}\right)=x^{2}+y^{2}$