Because \(n=5,\) our \(\large\binom{5}{r}\normalsize\) values will be the fifth row of Pascal's Triangle:

$$ \begin{gather}1\\ 1\quad 1\\ 1\quad 2\quad 1\\ 1\quad 3\quad 3\quad 1\\ 1\quad4\quad 6\quad4 \quad 1\\ 1\quad5\quad 10\quad10\quad5\quad 1 \end{gather} $$

Now we apply the binomial theorem:

$$ (2x-y)^5 = \sum^5_{r=0} \binom{5}{r}\ (2x)^{5-r}\ y^{r} $$

$$ \begin{eqnarray} &=& (2x)^5+5(2x)^{4}(-y)+10(2x)^{3}(-y)^2 \\[6pt] && \:+10(2x)^{2}(-y)^3+5(2x)(-y)^4+(-y)^5 \\[12pt] &=& 32x^5-80x^{4}y+80x^{3}y^2 \\[6pt] && \:-40x^{2}y^3+10xy^4-y^5 \end{eqnarray} $$

This is similar to the previous example, but care is needed over the powers of the fractions.

Because \(n=3,\) our \(\large\binom{3}{r}\normalsize\) values will be the third row of Pascal's Triangle:

$$ \begin{gather}1\\ 1\quad 1\\ 1\quad 2\quad 1\\ 1\quad 3\quad 3\quad 1\\ \end{gather} $$

Now we apply the binomial theorem:

$$ \left(\small\frac{2}{y^2}\normalsize-5y\right)^{\!3} = \sum^3_{r=0} \binom{3}{r}\left(\small\frac{2}{y^2}\normalsize\right)^{\!3-r}\ (-5y)^{r} $$

$$ \begin{eqnarray} &=& \binom{3}{0}\!\left(\small\!\frac{2}{y^2}\normalsize\!\right)^{\!3}+\binom{3}{1}\!\left(\small\!\frac{2}{y^2}\normalsize\!\right)^{\!2}\!(-5y)^1 \\[6pt] && +\binom{3}{2}\!\left(\!\small\frac{2}{y^2}\normalsize\!\right)^{\!1}\!(-5y)^2+\binom{3}{3}\!(-5y)^3 \\[12pt] &=& 1\left(\!\small\frac{8}{y^6}\normalsize\!\right)+3\left(\!\small\frac{4}{y^4}\normalsize\!\right)(-5y) \\[6pt] && +3\left(\!\small\frac{2}{y^2}\normalsize\!\right)(25y^2)+1(-125y^3) \\[12pt] &=& \small\frac{8}{y^6}\normalsize-\small\frac{60}{y^3}\normalsize+150-125y^3\\[6pt] \end{eqnarray} $$

The Advanced Higher Maths specification says that \(n\) won't be more than \(7\) if you have to generate the full expansion, but there is no upper limit on \(n\) if you are asked for the general term or a specific term.

(a)  When \(n=12,\) the general term is:

$$ \begin{eqnarray} && \binom{12}{r}\ \left(\frac{2}{x}\right)^{\!12-r}\ \left(-3x\right)^{r} \\[6pt] &=& \binom{12}{r}\ \left(\frac{2^{12-r}}{x^{12-r}}\right)\ \left(-3\right)^{r}x^{r} \\[6pt] &=& \binom{12}{r}\ \left(2^{12-r}\right)\ \left(-3\right)^{r}\ x^{r-(12-r)} \\[6pt] &=& \binom{12}{r}\ \left(2^{12-r}\right)\ \left(-3\right)^{r}\ x^{2r-12} \end{eqnarray} $$

(b)  First we find the value of \(r\) that corresponds to the term in \(x^8.\)

$$ \begin{eqnarray} 2r-12 &=& 8 \\[6pt] 2r &=& 20 \\[6pt] r &=& 10 \\[6pt] \end{eqnarray} $$

$$ \begin{eqnarray} && \binom{12}{r}\ \left(2^{12-r}\right)\ \left(-3\right)^{r}\ x^{2r-12} \\[6pt] &=& \binom{12}{10}\ \left(2^{12-10}\right)\ \left(-3\right)^{10}\ x^{8}\\[6pt] &=& \frac{12!}{10!(12-10)!} \ \left(2^{2}\right)\ \left(-3\right)^{10}\ x^{8}\\[6pt] &=& \frac{12\times 11}{2!} \ \left(2^{2}\right)\ \left(-3\right)^{10}\ x^{8}\\[6pt] &=& 66(4)(59049)x^{8}\\[6pt] &=& 15\,588\,936\,x^{8} \end{eqnarray} $$

So the coefficient of \(x^8\) is \(15\,588\,936.\)

(a)  When \(n=9,\) the general term is:

$$ \begin{eqnarray} && \binom{9}{r}\ \left(5x\right)^{9-r}\ \left(\frac{2}{x^2}\right)^{\!r}\\[6pt] &=& \binom{9}{r}\ \left(5^{9-r}\right)\ \left(x^{9-r}\right)\ \left(\frac{2^r}{x^{2r}}\right)\\[6pt] &=& \binom{9}{r}\ \left(5^{9-r}\right)\ \left(x^{9-r}\right)\ \left(2^r\right)\ \left(x^{-2r}\right)\\[6pt] &=& \binom{9}{r}\ \left(5^{9-r}\right)\ \left(2^r\right)\ x^{9-r-2r}\\[6pt] &=& \binom{9}{r}\ \left(5^{9-r}\right)\ \left(2^r\right)\ x^{9-3r}\\[6pt] \end{eqnarray} $$

(b)  First we find the value of \(r\) for the term that is independent of \(x,\) which is the \(x^0\) term.

$$ \begin{eqnarray} 9-3r &=& 0 \\[6pt] 3r &=& 9 \\[6pt] r &=& 3 \\[6pt] \end{eqnarray} $$

$$ \begin{eqnarray} && \binom{9}{r}\ \left(5^{9-r}\right)\ \left(2^r\right)\ x^{9-3r}\\[6pt] &=& \binom{9}{3}\ \left(5^{9-3}\right)\ \left(2^3\right)\ x^{0}\\[6pt] &=& \frac{9!}{3!\,(9-3)!}\ \left(5^6\right)\ \left(2^3\right)\\[6pt] &=& \frac{9!}{3!\,6!}\ \left(15\,625\right)\ \left(8\right)\\[6pt] &=& \frac{9\times 8\times 7}{3\times 2\times 1}\ \left(15\,625\right)\ \left(8\right)\\[6pt] &=& \left(84\right)\ \left(15\,625\right)\ \left(8\right)\\[6pt] &=& 10\,500\,000 \end{eqnarray} $$

(a)  When \(n=6,\) the general term is:

$$ \begin{eqnarray} && \binom{6}{r}\ \left(3x^2\right)^{6-r}\ \left(-\frac{a}{x^3}\right)^{\!r}\\[6pt] &=& \binom{6}{r}\ \left(3^{6-r}\right)\ \left(x^2\right)^{6-r}\ \left(-a\right)^r\ x^{-3r}\\[6pt] &=& \binom{6}{r}\ \left(3^{6-r}\right)\ \left(x^{12-2r}\right)\ \left(-a\right)^r\ x^{-3r}\\[6pt] &=& \binom{6}{r}\ \left(3^{6-r}\right)\ \left(-a\right)^r\ x^{12-2r-3r}\\[6pt] &=& \binom{6}{r}\ \left(3^{6-r}\right)\ \left(-a\right)^r\ x^{12-5r}\\[6pt] \end{eqnarray} $$

(b)  First we find the value of \(r\) that corresponds to the term in \(x^2.\)

$$ \begin{eqnarray} 12-5r &=& 2 \\[6pt] -5r &=& -\!10 \\[6pt] 5r &=& 10 \\[6pt] r &=& 2 \\[6pt] \end{eqnarray} $$

$$ \begin{eqnarray} && \binom{6}{r}\ \left(3^{6-r}\right)\ \left(-a\right)^r\ x^{12-5r}\\[6pt] &=& \binom{6}{2}\ \left(3^{6-2}\right)\ \left(-a\right)^2\ x^{2}\\[6pt] &=& \frac{6!}{2!\,(6-2)!}\ \left(3^4\right)\ \left(a^2\right)\ x^{2}\\[6pt] &=& \frac{6!}{2!\,4!}\ \left(81\right)\ a^2\,x^2\\[6pt] &=& \frac{6\times 5}{2}\ \left(81\right)\ a^2\,x^2\\[6pt] &=& \left(15\right)\,\left(81\right)\,a^2\,x^2\\[6pt] &=& 1215\,a^2\,x^2 \end{eqnarray} $$

Finally, we can solve for \(a\).

$$ \begin{eqnarray} 1215\,a^2 &=& 19\,440 \\[6pt] a^2 &=& \frac{19\,440}{1215} \\[6pt] a^2 &=& 16 \\[6pt] a &=& 4 \end{eqnarray} $$

Note: We should not include the negative square root of \(16\) as we were told in the question that \(a\gt 0.\)

A question of this type hasn't appeared on an Advanced Higher exam paper since 2009, but it still seems fair enough to us!

$$ \begin{eqnarray} && 1.02^{4} \\[6pt] &=& (1+0.02)^{4} \\[6pt] &=& \small\sum^4_{r=0}\binom{4}{r}\normalsize\,(1)^{4-r}\,(0.02)^{r} \\[6pt] &=& \small\sum^4_{r=0}\binom{4}{r}\normalsize\ (0.02)^{r} \\[10pt] &=& 1(0.02^0)+4(0.02^1)+6(0.02^2) \\[6pt] && \:\:+4(0.02^3)+1(0.02^4) \\[10pt] &=& 1+4(0.02)+6(0.0004) \\[6pt] && \:\:+4(0.000008)+0.00000016 \\[10pt] &=& 1+0.08+0.0024 \\[6pt] && \:\:+0.000032+0.00000016 \\[10pt] &=& 1.08243216 \end{eqnarray} $$

The general term in the binomial expansion of \(\left(2-x\right)^{5}\) is:

$$ \begin{eqnarray} && \binom{5}{r}\,2^{5-r}\,(-x)^r\\[6pt] &=& \binom{5}{r}\,2^{5-r}\,(-1)^r\ x^r \\[6pt] \end{eqnarray} $$

The \(x^3\) term is obtained by the \(1\) from \(\left(1+\large\frac{x}{4}\right)\) being multiplied by the \(x^3\) term from the expansion of \(\left(2-x\right)^5\) plus the \(\large\frac{x}{4}\) being multiplied by the \(x^2\) term from the expansion of \(\left(2-x\right)^{5}.\)

The \(x^3\) term in \(\left(2-x\right)^5\) is:

$$ \begin{eqnarray} && \binom{5}{3}\,\left(2^{5-3}\right)\,(-1)^3\,x^3\\[6pt] &=& \frac{5!}{3!\,2!}\ \left(2^2\right)\,(-1)\,x^3 \\[6pt] &=& \frac{5\times 4}{2}\,(4)\,(-1)\,x^3 \\[6pt] &=& -40x^3 \\[6pt] \end{eqnarray} $$

The \(x^2\) term in \(\left(2-x\right)^5\) is:

$$ \begin{eqnarray} && \binom{5}{2}\ 2^{5-2}\,(-1)^2\,x^2\\[6pt] &=& \frac{5!}{2!\,3!}\,\left(2^3\right)\,(1)\,x^2 \\[6pt] &=& (10)(8)x^2 \\[6pt] &=& 80x^2 \\[6pt] \end{eqnarray} $$

So the coefficient of the \(x^3\) term in the expansion of \(\left(1+\large\frac{x}{4}\normalsize\right)\left(2-x\right)^5\) is \((1)(-40)+(\frac{1}{4})(80)=-20.\)