Series, Progressions and the Binomial Expansion: Question 9

Worked solution

Step 1: Write down the coefficient of $x^2$

In the expansion of $(1+x)^n$, the term containing $x^2$ is $\binom{n}{2}x^2$, so the coefficient of $x^2$ is

$\binom{n}{2} = \frac{n(n-1)}{2}.$

Step 2: Set up and solve the equation

We are told this coefficient equals $45$:

$\frac{n(n-1)}{2} = 45 \implies n(n-1) = 90 \implies n^2 - n - 90 = 0.$

Using the quadratic formula:

$n = \frac{1 \pm \sqrt{1+360}}{2} = \frac{1\pm\sqrt{361}}{2} = \frac{1\pm19}{2}.$

This gives $n=10$ or $n=-9$. Since $n$ must be a positive integer, $n=10$.

Check: $\binom{10}{2} = \dfrac{10\times9}{2} = \dfrac{90}{2} = 45$ ✓, matching the given coefficient exactly.

Why the other options are wrong

• A ($9$): comes from correctly factorising $n(n-1)=90=9\times10$ but reporting the smaller factor instead of $n$ (the larger of the two consecutive integers).
• C ($19$): this is the value of $\sqrt{361}$ from the quadratic formula, mistakenly given as the final answer instead of completing the calculation $\dfrac{1+19}{2}=10$.
• D ($45$): this treats the coefficient of $x^1$ (which is simply $n$) as if it were the coefficient of $x^2$, giving $n=45$ directly without using $\binom{n}{2}$ at all.

Final answer

• $n = \boxed{10}$, option B.