Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $(1-2x)^5$ - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 2

To find the first four terms in the binomial expansion of $(1 - 2x)^5$, we apply the Binomial Theorem:

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

Where:

• $a = 1$
• $b = -2x$
• $n = 5$

Substituting these values yields:

1. For $k = 0$: $T_0 = \binom{5}{0}(1)^{5}(-2x)^{0} = 1$
2. For $k = 1$: $T_1 = \binom{5}{1}(1)^{4}(-2x)^{1} = -10x$
3. For $k = 2$: $T_2 = \binom{5}{2}(1)^{3}(-2x)^{2} = \binom{5}{2}(4x^2) = 60x^2$
4. For $k = 3$: $T_3 = \binom{5}{3}(1)^{2}(-2x)^{3} = \binom{5}{3}(-8x^3) = -80x^3$

Thus, the first 4 terms in ascending powers of $x$ are: $1 - 10x + 60x^2 - 80x^3$