Advanced Higher Maths
Binomial Theorem

\(n=5\small,\) so we need the values of \(\large\binom{5}{r}\small,\) which are in the row of Pascal's triangle that starts \(1\:5\small\ldots\)

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

Now we apply the binomial theorem:

$$ \begin{eqnarray} && (2x-y)^5 = \sum^5_{r=0} \binom{5}{r}\ (2x)^{5-r}\,(-y)^{r} \\[15pt] &=& (2x)^5+5(2x)^{4}(-y)+10(2x)^{3}(-y)^2 \\[6pt] && \:+10(2x)^{2}(-y)^3+5(2x)(-y)^4+(-y)^5 \\[18pt] &=& 32x^5-80x^{4}y+80x^{3}y^2 \\[6pt] && \:-40x^{2}y^3+10xy^4-y^5 \end{eqnarray} $$

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

\(n=3\small,\) so we need the values of \(\large\binom{3}{r}\small,\) which are in the row of Pascal's triangle that starts \(1\:3\small\ldots\)

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

Now we apply the binomial theorem:

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

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

The AH 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 \(n=12\small,\) the general term is:

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

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

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

Now we substitute \(r=10\) into the general term:

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

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

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

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

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

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

Now we substitute \(r=3\) into the general term:

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

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

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

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

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

Now we substitute \(r=2\) into the general term:

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

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

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

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

A question like this hasn't appeared on an Advanced Higher exam paper since 2009 Q8 , but it still seems fair enough to us!

The trick is to think of the \(1.02^4\) as \((1+0.02)^4\) and expand using the binomial theorem. Any power of \(1\) is just \(1\small,\) and powers of \(0.02\) are easy because it only has one significant figure.

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

The required \(x^3\) term consists of:

• the \(1\) from \(\left(1+\large\frac{x}{4}\right)\) multiplied by the \(x^3\) term from \(\left(2-x\vphantom{\large\frac{x}{4}\normalsize}\right)^5\)
• plus the \(\large\frac{x}{4}\) from \(\left(1+\large\frac{x}{4}\right)\) multiplied by the \(x^2\) term from \(\left(2-x\vphantom{\large\frac{x}{4}\normalsize}\right)^{5}\small.\)

So now we need to obtain the \(x^3\) and \(x^2\) terms from the expansion of \(\left(2-x\vphantom{\large\frac{x}{4}\normalsize}\right)^5\small.\)

The \(x^3\) term in \(\left(2-x\vphantom{\large\frac{x}{4}\normalsize}\right)^5\) is:

$$ \begin{eqnarray} && \binom{5}{3}\,2^{5-3}\,(-1)^3\,x^3\\[8pt] &=& \frac{5!}{3!\,2!}\,\left(2^2\right)\,(-1)\,x^3 \\[8pt] &=& (10)\,(4)\,(-1)\,x^3 \\[8pt] &=& -\!40x^3 \\[6pt] \end{eqnarray} $$

The \(x^2\) term in \(\left(2-x\vphantom{\large\frac{x}{4}\normalsize}\right)^5\) is:

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

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

(a)  When \(n=7\small,\) the general term is:

$$ \begin{eqnarray} && \binom{7}{r}\,\left(2x^2\right)^{7-r}\,\left(-\frac{d}{x^3}\right)^{\!r}\\[6pt] &=& \binom{7}{r}\,2^{7-r}\,\left(x^2\right)^{7-r}\,\left(-d\right)^r\,x^{-3r}\\[6pt] &=& \binom{7}{r}\,2^{7-r}\,x^{14-2r}\,\left(-d\right)^r\,x^{-3r}\\[6pt] &=& \binom{7}{r}\,2^{7-r}\,\left(-d\right)^r\,x^{14-2r-3r}\\[6pt] &=& \binom{7}{r}\,2^{7-r}\,\left(-d\right)^r\,x^{14-5r}\\[6pt] \end{eqnarray} $$

(b)  First we find the value of \(r\) that corresponds to the term in \(x^{-1}.\)

$$ \begin{gather} 14-5r=-1\\[6pt] 5r=15\\[6pt] r=3\\[6pt] \end{gather} $$

Now we can substitute \(r=3\) into the general coefficient and solve for \(d\):

$$ \begin{gather} \binom{7}{3}\,2^{7-3}\,\left(-d\right)^3=-70\,000\\[6pt] \frac{7!}{3!\,(7-3)!}\,\left(2^4\right)\,\left(-d\right)^3=-70\,000\\[6pt] \frac{7\!\times\!6\!\times\!5}{3\!\times\!2}\,\left(16\right)\left(d^3\right)=70\,000\\[6pt] \left(35\right)\left(16d^3\right)=70\,000\\[6pt] 560\,d^3=70\,000\\[6pt] d^3=\frac{70\,000}{560}\\[6pt] d^3=125 \\[6pt] d=5 \end{gather} $$

(a)  When \(n=8\small,\) the general term is:

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

(b)  First we find the value of \(r\) that corresponds to the term in \(x^{-1}\small.\)

$$ \begin{gather} 8-3r=-1 \\[6pt] 3r=9\\[6pt] r=3\\[6pt] \end{gather} $$

Now we substitute \(r=3\) into the general term:

$$ \begin{eqnarray} && \binom{8}{r}\,3^{8-r}\,\left(-2\right)^r\,x^{8-3r}\\[6pt] &=& \binom{8}{3}\,3^{8-3}\,\left(-2\right)^3\,x^{-1}\\[6pt] &=& \frac{8!}{3!\,5!}\,3^{5}\,\left(-8\right)\,x^{-1}\\[6pt] &=& \frac{8\!\times\!7\!\times\!6}{3!}\,\left(243\right)\left(-8\right)\,x^{-1}\\[6pt] &=& 56(243)(-8)x^{-1}\\[6pt] &=& -\!108\,864\,x^{-1} \end{eqnarray} $$

So the coefficient of \(x^8\) is \(-108\,864\small.\)

\(n=4\small,\) so we need the values of \(\large\binom{4}{r}\small,\) which are in the row of Pascal's triangle that starts \(1\:4\small\ldots\)

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

Now we apply the binomial theorem, and simplify:

$$ \begin{eqnarray} && \!\!\!\!\left(\frac{1}{x}-3x\right)^{4}\:=\:\sum^4_{r=0} \binom{4}{r}\,\left(\frac{1}{x}\right)^{\!4-r}(-3x)^{r}\\[18pt] &=& \binom{4}{0}\,\left(\!\frac{1}{x}\!\right)^{\!4}(-3x)^0+\binom{4}{1}\,\left(\!\frac{1}{x}\!\right)^{\!3}(-3x)^1 \\[6pt] && +\binom{4}{2}\,\left(\!\frac{1}{x}\!\right)^{\!2}(-3x)^2+\binom{4}{3}\,\left(\!\frac{1}{x}\!\right)^{\!1}(-3x)^3 \\[6pt] && +\binom{4}{4}\,\left(\!\frac{1}{x}\!\right)^{\!0}(-3x)^4 \\[18pt] &=& 1\left(\!\frac{1}{x^4}\!\right)+4\left(\!\frac{1}{x^3}\!\right)(-3x)+6\left(\!\frac{1}{x^2}\!\right)(9x^2) \\[6pt] && +4\left(\!\frac{1}{x}\!\right)(-27x^3)+1(81x^4) \\[18pt] &=& \frac{1}{x^4}-\frac{12}{x^2}+54-108x^2+81x^4\\[6pt] \end{eqnarray} $$