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Power Law

If 3x-14=a4x4...

Question

If ${(3 x - 1)}^{4} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x^{} + a_{0} .$ Find the value of $a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}$

A

$0$

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B

$- 1$

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C

$2$

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D

$- 2$

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Solution

The correct option is A
$0$

Explanation for the correct answer:
Finding the value of $a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}$
Given that ${(3 x - 1)}^{4} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
On Putting, $x = \frac{1}{3}$ then we have,
${(3 \times \frac{1}{3} - 1)}^{4} = a_{4} {(\frac{1}{3})}^{4} + a_{3} {(\frac{1}{3})}^{3} + a_{2} {(\frac{1}{3})}^{2} + a_{1} (\frac{1}{3}) + a_{0}$
⇒ $0 = \frac{a_{4}}{81} + \frac{a_{3}}{27} + \frac{a_{2}}{9} + \frac{a_{1}}{3} + a_{0}$
⇒ $0 = \frac{a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}}{81}$
⇒ $0 = a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}$
Hence the correct answer is (A).

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Similar questions

(3 x - 1)^{4} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}

, then find the value of

a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}

(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

{(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . a_{10} x^{10}

, then value of

{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}

x + 1

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

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