Author Topic: Binomial Theorem (Read 1601 times) Tweet Share

MinterMints · « **on:** June 06, 2018, 09:27:38 pm »

I need help with the following question:

Find the values of a and b if the
5th term of (ax + b)^10 is 13440x^6
and the 8th term is -960x^3.

Thanks in advance!

RuiAce · « **Reply #1 on:** June 06, 2018, 09:28:51 pm »

In general, the HSC will not give questions that assume the ordering of the terms. From which textbook was this so I have a reference?

MinterMints · « **Reply #2 on:** June 06, 2018, 09:46:14 pm »

Maths in Focus

RuiAce · « **Reply #3 on:** June 06, 2018, 10:08:13 pm »

$\text{The general term is }T_{k+1} = \binom{10}{k} (ax)^{10-k}b^k\\ \text{and we're told that }T_5 = 13440x^6\text{ and }T_8 = -960x^3$
$\text{i.e.}\\ \begin{align*}\binom{10}{4}(ax)^6 b^4 = 13440 x^6& \implies \binom{10}{4} a^6b^4 = 13440\\ \binom{10}{7} (ax)^3 b^7 = -960x^3&\implies \binom{10}{7}a^3b^7 = -960 \end{align*}$
$\text{This is now a pair of simultaneous equations.}\\ \text{We may square the second to obtain}\\ \binom{10}{7}^2 a^6 b^{14} = 960^2$
$\text{Dividing this with the first equation gives}\\ \begin{align*}\frac{\binom{10}7^2 a^6 b^{14}}{\binom{10}4 a^6 b^4} &= \frac{960^2}{13440}\\ \therefore b^{10} &= \frac{960^2}{13440} \times \frac{\binom{10}{4}}{\binom{10}{7}^2}\\ b&= 1\end{align*}$
$\text{Subbing back into the second: }\binom{10}{7}a^3 = -960 \implies a = -2$
Edit: $b = -1$ is also a solution. Under this case, $a = 2$

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