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प्रश्न
If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n2)=
पर्याय
a − ib
a2 − b2
a2 + b2
none of these
उत्तर
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:
\[2 \times 5 \times 10 . . . . \times (1 + n^2 ) = a^2 + b^2\]
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