Question

If $(2+i)(2+2i)(2+3i)\cdots \cdots (2+ni)=x+iy$, then $5\cdot 8\cdot 13\cdots \cdots (4+n^2)=$ _______

Answer 1

$\because (2+i)(2+2i)(2+3i)\dots (2+ni)=x+iy$

$\Rightarrow |(2+i)(2+2i)(2+3i)\dots (2+ni)|$

$=|x+iy|$

$\Rightarrow |(2+i)||(2+2i)||(2+3i)|\dots |(2+ni)|$

$=|x+iy|$

$\Rightarrow \sqrt{2^2+1^2}\times \sqrt{2^2+2^2}\times \sqrt{2^2+3^2}\dots \sqrt{2^2+n^2}$

$=\sqrt{x^2+y^2}$

$\Rightarrow \sqrt{5}\times \sqrt{8}\times \sqrt{13}\dots \sqrt{4+n^2}=\sqrt{x^2+y^2}$

$\Rightarrow \sqrt{(5\cdot 8\cdot 13\cdot \dots (4+n^2\left)\right)}=\sqrt{x^2+y^2}$

$\Rightarrow 5\cdot 8\cdot 13\dots (4+n^2)=x^2+y^2$