A-Level AQA Maths Mechanics Quantities, Units & Modelling: Find the coefficient of $x^2$ in

Find the coefficient of $x^2$ in the binomial expansion of $(2x - \frac{3}{x})^8$. - AQA - A-Level Maths Mechanics - Question 3 - 2020 - Paper 2

Question 3

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Find the coefficient of $x^2$ in the binomial expansion of $(2x - \frac{3}{x})^8$.

Worked Solution & Example Answer:Find the coefficient of $x^2$ in the binomial expansion of $(2x - \frac{3}{x})^8$. - AQA - A-Level Maths Mechanics - Question 3 - 2020 - Paper 2

Step 1

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Answer

To find the coefficient of $x^2$ in the expansion of $(2x - \frac{3}{x})^8$ , we can use the binomial theorem which states:

$(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$

In this case, let $a = 2x$ and $b = -\frac{3}{x}$ , with $n=8$ .

Step 2

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Answer

The general term in the expansion can be expressed as:

$T_k = {8 \choose k} (2x)^{8-k} \left(-\frac{3}{x}\right)^{k}$

This simplifies to:

$T_k = {8 \choose k} (2^{8-k})(-3^k) x^{8-k-k} = {8 \choose k} (2^{8-k})(-3^k) x^{8-2k}$

We want to find the term where the power of $x$ is 2, i.e., $8 - 2k = 2$ . This leads us to:

$8 - 2k = 2 \Rightarrow 2k = 6 \Rightarrow k = 3$

Step 3

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Answer

Now substituting $k=3$ back into the expression for the general term:

$T_3 = {8 \choose 3} (2^{8-3})(-3^3) x^2$

Calculating the coefficient:

$= {8 \choose 3} (2^5)(-27) = 56 \cdot 32 \cdot (-27) = -48384$

Step 4

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Answer

Thus, the coefficient of $x^2$ in the expansion of $(2x - \frac{3}{x})^8$ is -48384.