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Modulus of a Complex Number

Let 1 + 2x2...

Question

Let $(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{20} x^{20}$ Then, $3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3} + 3 a_{4} + 2 a_{5} + . . . 2 a_{19} + 3 a_{20}$ equals to:

A

\frac{{5.3}^{20} - 3}{2}

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B

\frac{{5.3}^{20} + 3}{2}

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C

\frac{{5.3}^{20} + 1}{2}

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D

\frac{{5.3}^{20} - 1}{2}

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Solution

The correct option is C $\frac{{5.3}^{20} + 1}{2}$
Given $:$ ${(1 + 2 x)}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . a_{20} x^{20}$
$∴ {(1 - 2 x)}^{20} = a_{0} - a_{1} x + a_{2} x^{2} + . . . . . . a_{20} x^{20}$
$\Rightarrow {(1 + 2 x)}^{20} + {(1 - 2 x)}^{20}$
$= 2 (a_{0} + a_{2} x^{2} + a_{4} x^{4} . . . . . . a_{20} x^{20})$
Take $x = 1$
$\Rightarrow 3^{20} + ({- 1)}^{20} = 2 (a_{0} + a_{2} . . . . . . a_{20})$
$\Rightarrow \frac{3}{2} \times {3^{20} + 1} = \frac{3}{2} \times 2 {a_{0} + a_{2} . . . . . . a_{20}}$
$\Rightarrow \frac{3}{2} {3^{20} + 1} = 3 {a_{0} + a_{2} . . . . . . a_{20}}$
$∴ {(1 + 2 x)}^{20} - {(1 - 2 x)}^{20}$
$= 2 {(a}_{1} x + a_{3} x^{3} + . . . . . . a_{19} x^{19})$
Take $x = 1$
$\Rightarrow {(1 + 2 x)}^{20} - {(1 - 2 x)}^{20} = 2 (a_{1} + a_{3} + a_{5} . . . . . . a_{19})$
$\Rightarrow 2 (a_{1} + a_{3} + . . . . . . a_{19}) = 3^{20} - 1^{20} = 3^{20} - 1$
To Find $3 a_{0} + 2 a_{1} + {3 a}_{2} + 2 a_{3} + . . . . . . .$
$= 3 (a_{0} + a_{2} + . . . . . . a_{20}) + 2 (a_{1} + a_{3} + . . . . . . a_{19})$
$= \frac{3}{2} {3^{20} + 1} + 3^{20} - 1$
$= 3^{20} \times \frac{5}{2} + \frac{1}{2}$
Hence, the answer is $\frac{{5.3}^{20} + 1}{2} .$

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