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Product Law

Let 1 +2x20...

Question

Let $(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . + a_{20} x^{20}$ .Then, $3 a_{0} + 2 a_{1} + 3 a_{2} + 2 a_{3}, + 3 a_{4} + 2 a_{2} + . . . . 2 a_{19} + 3 a_{20}$ 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 B $\frac{{5.3}^{20} + 1}{2}$
Consider the given equation.
$(1 + 2 x)^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . + a_{20} x^{20}$

Put $x = 1$ . Therefore,
$3^{20} = a_{0} + a_{1} + a_{2} + . . . . . . . . . + a_{20}$ ...... (1)

Put $x = - 1$ .
$1 = a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . + a_{20}$ ...... (2)

Substract both the equations.
$3^{20} - 1 = 2 (a_{1} + a_{3} . . . . . . . . . + a_{19})$ ....... (3)

Add both the equations.
$3^{20} + 1 = 2 (a_{0} + a_{2} . . . . . . . . . + a_{20})$ ....... (4)

Therefore,
$2 (a_{1} + a_{3} . . . . . . . . . + a_{19}) + 3 (a_{0} + a_{2} . . . . . . . . . + a_{20}) = (3^{20} - 1) + \frac{3}{2} (3^{20} + 1)$
$= \frac{{5.3}^{20} + 1}{2}$

Hence, this is the required value.

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