Find the coefficient of $x^2$ in the expansion of $(1 + 2x)^7$ Circle your answer. - AQA - A-Level Maths Pure - Question 2 - 2018 - Paper 2

To find the coefficient of $x^2$ in the expansion of $(1 + 2x)^7$, we can use the Binomial Theorem which states that:

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

In our case, let $a = 1$, $b = 2x$, and $n = 7$. We need to find the term where the exponent of $x$ is 2, which corresponds to $k = 2$.

Thus, substituting in the values, we get:

$T_k = \binom{7}{2} (1)^{7-2} (2x)^2$

Calculating the binomial coefficient: $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$

Now, substituting back into the term: $T_2 = 21 \cdot 1^5 \cdot (2x)^2 = 21 \cdot 4x^2 = 84x^2$

So the coefficient of $x^2$ is 84.