If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ n 2 ) is equal to Options: A . &nbsp a 2 - b 2 B . &nbsp a 2 + b 2 C . &nbsp √ a 2 + b 2 D . &nbsp √ a 2 − b 2

Answer: Option B : 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+ n 2 ) = a 2 + b 2

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

Question

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

Options:

A .

a^{2}

b^{2}

B .

a^{2}

b^{2}

C .

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

D .

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

Answer: Option B
:
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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