MCQ
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then 2.5.10.17...$(1+\text{n}^2)=$
- A$\text{a}-\text{ib}$
- B$\text{a}^2-\text{b}^2$
- C$\text{a}^2+\text{b}^2$
- Dnone of these
Solution:
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get,
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is:
[Hint: Let ABC be the equilateral triangle with vertex A (h, k) and let $\text{D}(\alpha,\beta)$ be the point on BC. Then $\frac{2\alpha+\text{h}}{3}=0=\frac{2\beta+\text{k}}{3}.$ Also $\alpha+\beta-2=0$ and $\frac{\text{k}-0}{\text{h}-0}\times(-1)=-1\Big].$
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