Question

If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n²)=

विकल्प

• a − ib

• a² − b²

• a² + b²

• none of these

Answer 1

a² + b²

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get,

\[\left| \left( 1 + i \right)\left( 1 + 2i \right)\left( 1 + 3i \right) . . . . . . \left( 1 + ni \right) \right| = a + ib\]

\[\left| \left( 1 + i \right)\left( 1 + 2i \right)\left( 1 + 3i \right) . . . . . . \left( 1 + ni \right) \right| \text { can be wriiten as } \left| \left( 1 + i \right) \right| \left| \left( 1 + 2i \right) \right| \left| \left( 1 + 3i \right) \right| . . . . \left| \left( 1 + ni \right) \right|\]

\[ \therefore \sqrt{1^2 + 1^2} \times \sqrt{1^2 + 2^2} \times \sqrt{1^2 + 3^2} . . . . \times \sqrt{1^2 + n^2} = \sqrt{a^2 + b^2}\]

\[\Rightarrow \sqrt{2} \times \sqrt{5} \times \sqrt{10} . . . . \times \sqrt{1 + n^2} = \sqrt{a^2 + b^2}\]

Squaring on both the sides, we get:

\[2 \times 5 \times 10 . . . . \times (1 + n^2 ) = a^2 + b^2\]