4 (a) Show that the first three terms, in descending powers of x, of the expansion of (2x - 3)^{10} are given by 1024x^{10} + px^9 + qx^8 where p and q are integers to be found - AQA - A-Level Maths Pure - Question 4 - 2021 - Paper 3

To expand

$(2x - 3)^{10},$

we use the Binomial Theorem, which states:

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

where in our case, $a = 2x$, $b = -3$, and $n = 10$. So, the terms for $k = 0, 1, 2$ will be:

1. For $k = 0$: ${10 \choose 0} (2x)^{10} (-3)^{0} = 1 \cdot (2x)^{10} \cdot 1 = 1024 x^{10}.$
2. For $k = 1$: ${10 \choose 1} (2x)^{9} (-3)^{1} = 10 \cdot (2x)^{9} \cdot (-3) = -30 \cdot 2^{9} x^{9} = -15360 x^{9}.$
3. For $k = 2$: ${10 \choose 2} (2x)^{8} (-3)^{2} = 45 \cdot (2x)^{8} \cdot 9 = 405 \cdot 2^{8} x^{8} = 103680 x^{8}.$

Combining these, we obtain:

$1024 x^{10} - 15360 x^{9} + 103680 x^{8}$.

Thus, we have: $p = -15360$ and $q = 103680$.