If 1- i 1-2 i 1-3 i -1- ni - a - ib- then 2-5 -10 - - -1- n 2 is equal toA- -a2-b2B- a2-b2C- a2-b2D- -a2-b2

The correct option is C we have (1+i)(1+2i)(1+3i)......(1+ni) = a+ib nbsp; nbsp; .......(i) ⇒ (1+i)(1+2i)(1+3i)......(1+ni) = a ib ........(ii) Multiplyin ...

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Byju's Answer

Standard XI

Mathematics

Algebra of Complex Numbers

If 1+ i 1+2 i...

Question

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ $n^{2}$ ) is equal to

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Solution

The correct option is B

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

$\Rightarrow$ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)

Multiplying (i) and (ii), we get 2.5.10.....(1+ $n^{2}$ ) = $a^{2}$ + $b^{2}$

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Q. If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+

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Q. If

(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2})

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\sqrt{a^{2} + b^{2}}

(b)

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a + b

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If (1 + i) (1 + 2i) (1 + 3i) ...... (1+ni) = a + ib, then 2.5.10.17 ...... ( $1 + n^{2}$ ) =

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If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+n2) is equal to

The correct option is B we have (1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i) ⇒ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii) Multiplying (i) and (ii), we get 2.5.10.....(1+n2) = a2+b2

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ $n^{2}$ ) is equal to

The correct option is B

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

$\Rightarrow$ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)

Multiplying (i) and (ii), we get 2.5.10.....(1+ $n^{2}$ ) = $a^{2}$ + $b^{2}$