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 $(1+i)(1+2i)(1+3i)......(1+ni)$
$=x+iy\dots (1)$
$\therefore (1-i)(1-2i)(1-3i)....(1-ni)$
$=x-iy\dots (2)$
Multilying $(1)$ and $(2)$ we get,
$\left(1-i^2\right)\left(1-4i^2\right)\left(1-9i^2\right).....\left(1-n^2{i}^2\right)$
$=x^2-i^2{y}^2$
$\Rightarrow \left(1+1\right)\left(1+4\right)\left(1+9\right)....\left(1+n^2\right)=x^2+y^2$
$\Rightarrow 2\cdot 5\cdot \mathrm{10.....}+\left(1+n^2\right)=x^2+y^2$