Unknown Binomial Coefficients | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Turn a coefficient condition into an equation

A coefficient may contain an unknown constant or the unknown power $n$ . Write just the coefficient being described, then translate “equals” or “twice” into an equation.

Worked example

The coefficient of $x^2$ in $(1+x/2)^n$ is 7. Find the positive integer $n$ .

A non-zero $x^2$ coefficient requires $n\ge2$ . With $n$ unknown, use the factorial expression rather than a numerical nCr entry. Expand only as far as the shared factor:

n!=n(n-1)(n-2)!

\begin{gathered}\binom n2=\frac{n!}{2!(n-2)!}\\=\frac{n(n-1)(n-2)!}{2!(n-2)!}\\=\frac{n(n-1)}2\end{gathered}

The shared $(n-2)!$ is non-zero, so it cancels. Also, $2!=2$ .

\frac{n(n-1)}2\left(\frac12\right)^2=7

n^2-n-56=0

The polynomial solver gives 8 and −7, so write $(n-8)(n+7)=0$ . Only $n=8$ is a positive integer. Check: $\binom82/4=28/4=7$ .

Keep zero and both signs until checked

In $(1+px)^4$ , the coefficients of $x$ and $x^2$ are $4p$ and $6p^2$ . If they are equal, factor instead of immediately dividing by $p$ :

6p^2=4p\quad\Rightarrow\quad 2p(3p-2)=0

Both $p=0$ and $p=2/3$ work unless a condition excludes zero. A squared coefficient can likewise give two signs.

Use the polynomial solver first for an ordinary quadratic. Convenient roots suggest factorised working. For awkward roots, use the quadratic formula to keep exact working. For example, $p^2+p-1=0$ gives $p=(-1\pm\sqrt5)/2$ , not just two rounded decimals.

Now you try

Your turn9709/11 May/Jun 2025 Q5 [7 marks]

(a) Find the first three terms, in ascending powers of $x$ , of $(2-px)^5$ and $(1-x/2)^4$ .

(b) The coefficient of $x^2$ in their product is 93. Find the possible values of $p$ .

Show worked answer

(a) The requested terms are

32-80px+80p^2x^2

1-2x+\frac32x^2

(b) Include all three power pairs:

\begin{aligned}32\left(\frac32\right)&+(-80p)(-2)\\&+80p^2=93\end{aligned}

80p^2+160p-45=0

The solver gives $p=1/4,-9/4$ . Write the factorisation:

5(4p-1)(4p+9)=0

Answer

p=\frac14\quad\text{or}\quad p=-\frac94

Both values give coefficient 93. No positive-only condition was given.

Your turnIndependent practice

(a) The coefficient of $x^2$ in $(1-x/3)^n$ is $5/3$ . Find the positive integer $n$ .

(b) In $(1+px)^5$ , the coefficient of $x^2$ is twice the coefficient of $x$ . Find all values of $p$ .

Show worked answer

(a) The square makes the coefficient positive:

\frac{n(n-1)}2\left(-\frac13\right)^2=\frac53

(n-6)(n+5)=0

Keep $n=6$ ; −5 is not a positive integer. Check: $\binom62/9=15/9=5/3$ .

(b) $10p^2=2(5p)$ , so $10p(p-1)=0$ . Both $p=0$ and $p=1$ satisfy the original condition.

Turn a coefficient condition into an equation

A coefficient may contain an unknown constant or the unknown power $n$ . Write just the coefficient being described, then translate “equals” or “twice” into an equation.

Worked example

The coefficient of $x^2$ in $(1+x/2)^n$ is 7. Find the positive integer $n$ .

A non-zero $x^2$ coefficient requires $n\ge2$ . With $n$ unknown, use the factorial expression rather than a numerical nCr entry. Expand only as far as the shared factor:

n!=n(n-1)(n-2)!

\begin{gathered}\binom n2=\frac{n!}{2!(n-2)!}\\=\frac{n(n-1)(n-2)!}{2!(n-2)!}\\=\frac{n(n-1)}2\end{gathered}

The shared $(n-2)!$ is non-zero, so it cancels. Also, $2!=2$ .

\frac{n(n-1)}2\left(\frac12\right)^2=7

n^2-n-56=0

The polynomial solver gives 8 and −7, so write $(n-8)(n+7)=0$ . Only $n=8$ is a positive integer. Check: $\binom82/4=28/4=7$ .

Keep zero and both signs until checked

In $(1+px)^4$ , the coefficients of $x$ and $x^2$ are $4p$ and $6p^2$ . If they are equal, factor instead of immediately dividing by $p$ :

6p^2=4p\quad\Rightarrow\quad 2p(3p-2)=0

Both $p=0$ and $p=2/3$ work unless a condition excludes zero. A squared coefficient can likewise give two signs.

Now you try

Your turn9709/11 May/Jun 2025 Q5 [7 marks]

(a) Find the first three terms, in ascending powers of $x$ , of $(2-px)^5$ and $(1-x/2)^4$ .

(b) The coefficient of $x^2$ in their product is 93. Find the possible values of $p$ .

Show worked answer

(a) The requested terms are

32-80px+80p^2x^2

1-2x+\frac32x^2

(b) Include all three power pairs:

\begin{aligned}32\left(\frac32\right)&+(-80p)(-2)\\&+80p^2=93\end{aligned}

80p^2+160p-45=0

The solver gives $p=1/4,-9/4$ . Write the factorisation:

5(4p-1)(4p+9)=0

Answer

p=\frac14\quad\text{or}\quad p=-\frac94

Both values give coefficient 93. No positive-only condition was given.

Your turnIndependent practice

(a) The coefficient of $x^2$ in $(1-x/3)^n$ is $5/3$ . Find the positive integer $n$ .

(b) In $(1+px)^5$ , the coefficient of $x^2$ is twice the coefficient of $x$ . Find all values of $p$ .

Show worked answer

(a) The square makes the coefficient positive:

\frac{n(n-1)}2\left(-\frac13\right)^2=\frac53

(n-6)(n+5)=0

Keep $n=6$ ; −5 is not a positive integer. Check: $\binom62/9=15/9=5/3$ .

(b) $10p^2=2(5p)$ , so $10p(p-1)=0$ . Both $p=0$ and $p=1$ satisfy the original condition.