If (1 + i)(1 + 2i)(1 + 3i) dots (1 + ni) = a | vedclass

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(B) Given that $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$ ..... $(i)$Taking the conjugate on both sides,we get $(1 - i)(1 - 2i)(1 - 3i) \dots (

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$a^2 + b^2$

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(B) Given that $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$ ..... $(i)$Taking the conjugate on both sides,we get $(1 - i)(1 - 2i)(1 - 3i) \dots (1 - ni) = a - ib$ ..... $(ii)$Multiplying equation $(i)$ and equation $(ii)$,we get:$[(1 + i)(1 - i)] 	imes [(1 + 2i)(1 - 2i)] 	imes \dots 	imes [(1 + ni)(1 - ni)] = (a + ib)(a - ib)$Since $(1 + ki)(1 - ki) = 1^2 - (ki)^2 = 1 + k^2$,the expression becomes:$(1 + 1^2)(1 + 2^2)(1 + 3^2) \dots (1 + n^2) = a^2 + b^2$$2 	imes 5 	imes 10 	imes \dots 	imes (1 + n^2) = a^2 + b^2$

If $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$,then $2 \times 5 \times 10 \times \dots \times (1 + n^2)$ is equal to

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