If (1 + i) cdot (1 + 2i) dots dots dots (1 + | vedclass

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(A) Given: $(1 + i)(1 + 2i) \dots (1 + ni) = x + iy$. Taking the modulus on both sides: $|(1 + i)(1 + 2i) \dots (1 + ni)| = |x + iy|$. Using the prope

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$x^2 + y^2$

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(A) Given: $(1 + i)(1 + 2i) \dots (1 + ni) = x + iy$. Taking the modulus on both sides: $|(1 + i)(1 + 2i) \dots (1 + ni)| = |x + iy|$. Using the property $|z_1 z_2 \dots z_n| = |z_1| |z_2| \dots |z_n|$, we get: $|1 + i| |1 + 2i| \dots |1 + ni| = |x + iy|$. Since $|a + bi| = \sqrt{a^2 + b^2}$, we have: $\sqrt{1^2 + 1^2} \cdot \sqrt{1^2 + 2^2} \cdot \dots \cdot \sqrt{1^2 + n^2} = \sqrt{x^2 + y^2}$. This simplifies to: $\sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \dots \sqrt{1 + n^2} = \sqrt{x^2 + y^2}$. Squaring both sides: $2 \cdot 5 \cdot 10 \dots (1 + n^2) = x^2 + y^2$.

If $(1 + i) \cdot (1 + 2i) \dots \dots \dots (1 + ni) = x + iy$ (where $i = \sqrt{-1}$), then find the value of $(2) \cdot (5) \cdot (10) \dots \dots \dots (1 + n^2)$.

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