Series, Progressions and the Binomial Expansion: Question 4

Worked solution

Part (a): Expanding $(1-2x)^6$

Using the binomial expansion with $a=1$ and $b=-2x$:

$(1-2x)^6 = \sum_{r=0}^{6}\binom{6}{r}(1)^{6-r}(-2x)^r = \sum_{r=0}^{6}\binom{6}{r}(-2)^r x^r.$

Computing each coefficient up to $r=3$:

• $r=0$: $\binom{6}{0}(-2)^0 = 1\times1 = 1$
• $r=1$: $\binom{6}{1}(-2)^1 = 6\times(-2) = -12$
• $r=2$: $\binom{6}{2}(-2)^2 = 15\times4 = 60$
• $r=3$: $\binom{6}{3}(-2)^3 = 20\times(-8) = -160$

Check: using Pascal’s triangle, row $6$ is $1,6,15,20,15,6,1$, and multiplying by $(-2)^r=1,-2,4,-8,\ldots$ for $r=0,1,2,3$ gives the same values $1, -12, 60, -160$.

So

$(1-2x)^6 = 1 - 12x + 60x^2 - 160x^3 + \ldots$

Part (b): Coefficient of $x^3$ in $(2+5x)(1-2x)^6$

Using the expansion from part (a), the coefficient of $x^3$ in the product $(2+5x)(1-12x+60x^2-160x^3+\ldots)$ comes from two contributions:

• $2 \times (\text{coefficient of } x^3) = 2\times(-160) = -320$
• $5x \times (\text{coefficient of } x^2) = 5\times60 = 300$ (since $5x\times60x^2 = 300x^3$)

Adding these:

$-320 + 300 = -20.$

Check by expanding the product term by term up to $x^3$:

$(2+5x)(1-12x+60x^2-160x^3) = 2 - 24x + 120x^2 - 320x^3 + 5x - 60x^2 + 300x^3 + \ldots$

Collecting the $x^3$ terms: $-320x^3+300x^3 = -20x^3$, confirming the coefficient is $-20$.

Final answers

• (a) $(1-2x)^6 = 1 - 12x + 60x^2 - 160x^3 + \ldots$
• (b) Coefficient of $x^3$ in $(2+5x)(1-2x)^6$ is $\boxed{-20}$.