Binomial Expansion: Finding an Unknown Constant

The efficient way to hit one coefficient in a binomial expansion is the general term, not a full expansion. For $(a + b)^n$ the term in $b^r$ is $\binom{n}{r}a^{n-r}b^r$. Choose the $r$ that produces the power you need, and everything else drops away.

In the expansion of $(2 + kx)^6$, the coefficient of $x^2$ is $60$. Find the possible values of the constant $k$. [5]

The working

Step 1, write the general term for $(2 + kx)^6$, taking $a = 2$, $b = kx$: $\binom{6}{r}(2)^{6-r}(kx)^r$

Step 2, choose $r$ for the $x^2$ term. The power of $x$ is $r$, so set $r = 2$: $\binom{6}{2}(2)^{4}(kx)^2 = 15 \times 16 \times k^2 x^2 \quad \text{(M1 for } \binom{6}{2}\text{ and powers)}$

Step 3, extract the coefficient (everything except $x^2$): $\text{coefficient of } x^2 = 15 \times 16 \times k^2 = 240k^2 \quad \text{(A1)}$

Step 4, set equal to $60$ and solve: $240k^2 = 60 \;\Rightarrow\; k^2 = \frac{60}{240} = \frac{1}{4} \;\Rightarrow\; k = \pm\frac{1}{2} \quad \text{(M1, A1)}$

Where the marks are won and lost

• $\binom{6}{2} = 15$ and $2^4 = 16$ must both be included. Forgetting the $2^{4}$ (from the $a^{n-r}$ part) is the most common error, it changes the coefficient completely.
• $(kx)^2 = k^2 x^2$: the constant $k$ is squared too. Writing $k$ instead of $k^2$ gives a linear equation and the wrong answer.
• $k^2 = \frac14$ has two roots, $k = \pm\frac12$. Unless the question restricts the sign, give both.

Common mistakes

• Using $2^6$ or $2^2$ instead of $2^{6-2} = 2^4$.
• Forgetting to square $k$ inside $(kx)^2$.
• Giving only $k = \frac12$ and dropping the negative root.

Full method: Binomial Expansion notes. Topic home: Series pillar.