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In the binomial expansion of $(a + 2x)^{7}$ where $a$ is a constant the coefficient of $x^{4}$ is 15120 Find the value of $a.$ - Edexcel - A-Level Maths Pure - Question 6 - 2020 - Paper 2

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In-the-binomial-expansion-of--$(a-+-2x)^{7}$-where-$a$-is-a-constant--the-coefficient-of-$x^{4}$-is-15120--Find-the-value-of-$a.$-Edexcel-A-Level Maths Pure-Question 6-2020-Paper 2.png

In the binomial expansion of $(a + 2x)^{7}$ where $a$ is a constant the coefficient of $x^{4}$ is 15120 Find the value of $a.$

Worked Solution & Example Answer:In the binomial expansion of $(a + 2x)^{7}$ where $a$ is a constant the coefficient of $x^{4}$ is 15120 Find the value of $a.$ - Edexcel - A-Level Maths Pure - Question 6 - 2020 - Paper 2

Step 1

Find the coefficient of $x^{4}$ in the expansion

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Answer

The general term in the binomial expansion of (a+2x)7(a + 2x)^{7} is given by:

Tk=(nr)anr(2x)rT_k = \binom{n}{r} a^{n-r} (2x)^r

where n=7n = 7 and r=4r = 4. So,

T4=(74)a74(2x)4=(74)a3(24x4)T_4 = \binom{7}{4} a^{7-4} (2x)^4 = \binom{7}{4} a^{3} (2^4 x^4)

Calculating \binom{7}{4}$, we have:

(74)=7!4!(74)!=7!4!3!=35\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35

Thus, the term becomes:

T4=35a3(16x4)=560a3x4T_4 = 35 a^{3} (16 x^4) = 560 a^{3} x^4

Step 2

Set the coefficient equal to 15120

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Answer

To find the value of aa, we set the coefficient of x4x^4 from our term equal to 15120:

560a3=15120560 a^{3} = 15120

Dividing both sides by 560 gives:

a3=15120560=27a^{3} = \frac{15120}{560} = 27

Step 3

Solve for $a$

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Answer

To find aa, we take the cube root of both sides:

a=273=3a = \sqrt[3]{27} = 3

Therefore, the value of aa is 3.

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