Series, Progressions and the Binomial Expansion: Question 6

Worked solution

Step 1: Write down the general term

For $\left(x - \dfrac{2}{x^2}\right)^6$, treat $a=x$ and $b=-\dfrac{2}{x^2}$. The general term is

$T_{r+1} = \binom{6}{r}\, x^{6-r}\left(-\frac{2}{x^2}\right)^r = \binom{6}{r}(-2)^r\, x^{6-r}\,x^{-2r} = \binom{6}{r}(-2)^r\, x^{6-3r}.$

Step 2: Find the value of $r$ that gives a constant term

The term is independent of $x$ when the power of $x$ is zero:

$6 - 3r = 0 \implies r = 2.$

Step 3: Compute the term

$\binom{6}{2} = \frac{6!}{2!\,4!} = \frac{720}{2\times24} = 15$

$(-2)^2 = 4$

$\text{constant term} = 15\times4 = 60.$

Check: listing every term of the expansion by increasing $r$ confirms this:

$r=0:\ x^6,\quad r=1:\ -12x^3,\quad r=2:\ 60,\quad r=3:\ -160x^{-3},\quad r=4:\ 240x^{-6},\quad r=5:\ -192x^{-9},\quad r=6:\ 64x^{-12}.$

Only the $r=2$ term has power $x^0$, and it equals $60$, confirming the result.

Why the other options are wrong

• A ($-60$): comes from computing $(-2)^2$ as $-4$ instead of $4$ (an order-of-operations error), then $15\times(-4)=-60$.
• C ($-160$): this is the actual $r=3$ term of the expansion, $\binom{6}{3}(-2)^3x^{-3}=-160x^{-3}$. A genuine term, but its power of $x$ is $-3$, not $0$, so it is not the constant term.
• D ($240$): this is the actual $r=4$ term, $\binom{6}{4}(-2)^4x^{-6}=240x^{-6}$, again a real term of the expansion, but with power $x^{-6}$, not $x^0$.

Final answer

• The term independent of $x$ is $\boxed{60}$, option B.