Series, Progressions and the Binomial Expansion: Question 1

Worked solution

Step 1: Write down the general term

For $(1+2x)^6$, the term containing $x^r$ in the expansion is

$T_{r+1} = \binom{6}{r} (1)^{6-r} (2x)^r = \binom{6}{r}\, 2^r x^r.$

Step 2: Choose $r$ so that the power of $x$ is $3$

We need $x^r = x^3$, so $r = 3$ (not the “4th term” counted from $1$. The power of $x$ and the value of $r$ are the same number here).

Step 3: Compute the coefficient

$\binom{6}{3} = \frac{6!}{3!\,3!} = \frac{720}{6 \times 6} = 20$

$2^3 = 8$

$\text{coefficient} = 20 \times 8 = 160$

Check: building up the expansion term by term confirms this ($r=0$ gives $1$, $r=1$ gives $\binom{6}{1}2^1 = 12$, $r=2$ gives $\binom{6}{2}2^2 = 60$, and $r=3$ gives $\binom{6}{3}2^3 = 160$, so $(1+2x)^6 = 1 + 12x + 60x^2 + 160x^3 + \ldots$) the coefficient of $x^3$ is indeed $160$.

Why the other options are wrong

• A ($60$): this is the coefficient of $x^2$ (using $r=2$), not $x^3$.
• C ($240$): comes from using $r=4$ instead of $r=3$: $\binom{6}{4}2^4 = 15 \times 16 = 240$.
• D ($320$): keeps the correct $\binom{6}{3}=20$ but uses the wrong power of $2$: $20 \times 2^4 = 320$ instead of $20\times2^3$.

Final answer

• Coefficient of $x^3$ $= \boxed{160}$, option B.