This question was extracted from Nan Hua High School Sec 4 A Maths Mid year Exam 2018.

Remember the use of algebraic expansion in sec 2 whereby \((a+b)^2\) where most students will just expand it as \((a+b)(a+b)\). The truth is that students are expected to memorise \((a\pm b)^2 = a^2 \pm 2ab + b^2\) to have an edge over others as well as to shorten the presentation in examinations.

This is the bare minimum that students have to know before they move on to “Binomial Expansions” in sec 3 Additiional Mathematics.

In Binomial expansions, when the powers of binomials get unrealistically big, it is simply cumbersome to work out the slow expansions. Binomial Theorem then sets in where students can observe a guided pattern upon expansion.

To do well in this chapter, besides the Sec 2 foundation in ALGEBRA and Sec 3 INDICES, students need to be well versed in the observation and the techniques that can be applied to the different types of questions in Binomial Expansions.

For students who first join Math Academy and requested us to check their Amath script, it is common to see a handful of students who will just expand slowly in a desperate bid just to score that few marks in examinations. Hmm, did they know that they are eating into the precious time of other questions? Some will eventually find that needle in the haystack, but some will give up after a few attempts of algebraic errors. Well, there are always more efficient ways of getting things done.

Let me highlight using the example extracted from 2018 Nan Hua High Paper.

Q1. (i) By considering the **general term** in the binomial expansion of \(\left(x^2 + \frac{2}{x} \right)^{15}\), explain why all of the **powers of ** \(x\) in this expansion are **multiples of 3**. [3]

(ii) Find the coefficient of \(x^{22}\) in the expansion of \(\left(3x + \frac{1}{x^2} \right)\left(x^2 + \frac{2}{x} \right)^{15}\). [4]

(i)

Let us apply the general term formula: \(T_{r+1} = ^{15}C_r(x^2)^{15-r}\left(\frac{2}{x} \right)^r\)

We are only keen to know the powers of \(x\), hence we will pull out the terms that contain \(x\) as the base.

\((x^{30-2r})\left(\frac{1}{x} \right)^r\)

\(= \frac{x^{30-2r}}{x^r}\)

\(=x^{30-3r}\)

\(=x^{3(10-r)}\)

Since the power of \(x\) is \(3(10-r)\), which is a multiple of 3, therefore all of the **powers of ** \(x\) in this expansion are **multiples of 3.**

(ii)

For expansions of 2 or more brackets, most students will just expand blindly without further thought. That is a waste of precious time in examination. This is a 4 marks question and we have to complete this within a 6 minutes timeframe.

We have to **first observe the combination** that will pair up to get \(x^{22}\).

\(3x\) in the first bracket needs to pair with \(x^{21}\) from the second bracket and \(\frac{1}{x^2}\) in the first bracket needs to pair with \(x^{24}\) from the second bracket.

That is, \(\left(3x + \frac{1}{x^2} \right)\left( x^2 + \frac{2}{x}\right)^{15} =\left(3x + \frac{1}{x^2} \right)\left(Ax^{21} + Bx^{24} + \ldots \right)\)

We attempt to retrieve only \(x^{21}\) and \(x^{24}\).

From the general term in part (i),

\(x^{30-3x} = x^{21}\)

\(r=3\)

Substitute \(r=3\) into general term,

\(T_{3+1} = T_4\)

\(=^{15}C_3(2)^3x^{21}\)

\(=3640x^{21}\)

and

\(x^{30-3x} = x^{24}\)

\(r=2\)

Substitute \(r=2\) into general term,

\(T_{2+1} = T_3\)

\(=^{15}C_2(2)^2x^{24}\)

\(=420x^{24}\)

The expansion then becomes \(\left(3x + \frac{1}{x^2} \right)\left(3640x^{21} + 420x^{24} + \ldots \right)\)

Coefficient of \(x^{22} = (3 \times 3640) + (1 \times 420) = 11340\)