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O-Level A-Math · 2020 · P2 Q3 Binomial · General term ⁿCr 6 marks: 3 + 3 · algebra (binomial theorem) difficulty 4 of 5

O-Level A-Math 2020 Paper 2, Question 3: General term ⁿCr

The answer

(i) explained (no integer \(r\) gives \(x^0\))
(ii) \(-3024\)

O-Level A-Math 2020 Paper 2 Question 3 · Verified worked solution by the Genius Plus Academy teaching team

The question

(i) By considering the general term in the binomial expansion of \(\left(\dfrac{3}{x^2} + x\right)^8\), explain why every term is dependent on \(x\). [3]

(ii) Find the term independent of \(x\) in the expansion of \(\left(\dfrac{3}{x^2} + x\right)^8 (5 - 2x)\). [3]

Step-by-step solution

(i) The general term is \[T_{r+1} = \binom{8}{r}\left(\dfrac{3}{x^2}\right)^{8-r}(x)^r = \binom{8}{r}\,3^{8-r}\,x^{-2(8-r)+r} = \binom{8}{r}\,3^{8-r}\,x^{3r-16}, \qquad r = 0, 1, \ldots, 8.\] A term independent of \(x\) needs \(3r - 16 = 0 \Rightarrow r = \dfrac{16}{3}\), which is not an integer in \(\{0, \ldots, 8\}\). So no term has \(x^0\), every term depends on \(x\).

(ii) In \(\left(\dfrac{3}{x^2} + x\right)^8 (5 - 2x)\), a constant term comes from \(5 \times (x^0\text{ term})\) and from \(-2x \times (x^{-1}\text{ term})\). As shown, there is no \(x^0\) term, so only the second contributes. For the \(x^{-1}\) term: \(3r - 16 = -1 \Rightarrow r = 5\), giving coefficient \(\binom{8}{5}3^{3} = 56 \times 27 = 1512\). Hence the term independent of \(x\) is \[(-2x)\big(1512\,x^{-1}\big) = -3024.\]

Answer: (i) explained (no integer \(r\) gives \(x^0\))
(ii) \(-3024\)

Same structure, different numbers

A question is hard because of its structure, not its surface.

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That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.

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What does O-Level A-Math 2020 Paper 2 Question 3 test?

It is a general term ⁿcr question from Binomial, worth 6 marks: 3 + 3.

Is this the same as IP Math?

Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.

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