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 Mechanics - Question 4 - 2021 - Paper 3

To find the first three terms of the expansion of (2x - 3)^{10}, we can use the Binomial Theorem:

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

Here, let:

• $a = 2x$
• $b = -3$
• $n = 10$

The first term (for k=0) is: ${10 \choose 0} (2x)^{10} (-3)^{0} = 1024x^{10}$

The second term (for k=1) is: ${10 \choose 1} (2x)^{9} (-3)^{1} = 10 \cdot 2^{9} x^{9} \cdot (-3) = -15360x^{9}$

The third term (for k=2) is: ${10 \choose 2} (2x)^{8} (-3)^{2} = 45 \cdot 2^{8} x^{8} \cdot 9 = 103680x^{8}$

Thus, the first three terms in descending powers of x are:

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

Therefore, we can identify: p = -15360 q = 103680.