If $(1-\mathrm{i})(1-2\mathrm{i})\dots \dots \dots (1-\mathrm{ni})=a-ib\text{, then }2.5.10\dots .\left(1+{\mathrm{n}}^2\right)=$

$$\begin{gathered} Given : (1-\mathrm{i})(1-2\mathrm{i})\dots \dots \dots (1-\mathrm{ni})=a+ib \\ Taking modulus on both sides of equation , we have \\ \left|(1-\mathrm{i})(1-2\mathrm{i})\dots \dots \dots (1-\mathrm{ni})\right|=\left|a+ib\right| \\ \Rightarrow \left|(1-\mathrm{i})\right|\left|(1-2\mathrm{i})\right|\dots \dots \dots \left|(1-\mathrm{ni})\right|=\left|a+ib\right| \\ \Rightarrow \sqrt{1^2+1^2}\times \sqrt{1^2+2^2}\times \sqrt{1^2+3^2}......\sqrt{1^2+n^2}=\sqrt{a^2+b^2} \\ \Rightarrow \sqrt{2}\times \sqrt{5}\times \sqrt{10}........\sqrt{1+n^2}=\sqrt{a^2+b^2} \\ On squaring on both sides , we have \\ \Rightarrow 2\times 5\times 10\times .......\times \left(1+n^2\right)=a^2+b^2 \end{gathered}$$