Unit 4 Review Exercises

The Binomial Theorem

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. Questions marked AHL are for HL students.

1 Paper 1 Evaluate \(\dbinom{10}{3}\) without a calculator.

Solution

\(\dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\). See 4.1.

2 Paper 1 Find the coefficient of \(x^3\) in the expansion of \((2x - 1)^5\).

Solution

The term in \(x^3\) is \(\dbinom{5}{2}(2x)^3(-1)^2 = 10 \times 8x^3\). Coefficient \(80\). See 4.2.

3 Paper 1 Find the constant term in the expansion of \(\left(x - \dfrac{2}{x}\right)^8\).

Solution

\(\dbinom{8}{r}x^{8-r}(-2)^r x^{-r} = \dbinom{8}{r}(-2)^r x^{8 - 2r}\). \(\;r = 4\): \(\;70 \times 16 = 1120\). See 4.2.

4 Paper 1 The coefficient of \(x^2\) in the expansion of \((1 + kx)^7\) is \(84\). Find the possible values of \(k\).

Solution

\(\dbinom{7}{2}k^2 = 21k^2 = 84 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2\). See 4.1.

5 Paper 1 Find the expansion of \((1 + x)^5(1 - 2x)^4\) up to the term in \(x^2\).

Solution

\((1 + 5x + 10x^2)(1 - 8x + 24x^2) = 1 - 8x + 24x^2 + 5x - 40x^2 + 10x^2 + \cdots = 1 - 3x - 6x^2 + \cdots\) See 4.1.

6 Paper 1 Find the coefficient of \(x^4\) in the expansion of \(\left(x^2 + \dfrac{3}{x}\right)^5\).

Solution

\(\dbinom{5}{r}x^{10 - 2r}3^r x^{-r} = \dbinom{5}{r}3^r x^{10 - 3r}\). \(\;10 - 3r = 4 \Rightarrow r = 2\): \(\;10 \times 9 = 90\). See 4.2.

7 Paper 2 (a) Write down the first three terms of \((1 + 2x)^{10}\).

(b) Use them to estimate \(1.002^{10}\).

Solution

(a) \(1 + 20x + 180x^2\).

(b) \(x = 0.001\): \(1 + 0.02 + 0.000\,18 = 1.020\,18\). See 4.1.

8 Paper 1 AHL Expand \((1 - 2x)^{-\frac{1}{2}}\) up to the term in \(x^2\), and state the interval of validity.

Solution

\(1 + \left(-\frac{1}{2}\right)(-2x) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-2x)^2 = 1 + x + \dfrac{3}{2}x^2 + \cdots\), \(\;|x| < \dfrac{1}{2}\). See 4.3.

9 Paper 1 AHL Expand \(\sqrt{9 - x}\) up to the term in \(x^2\), and state the interval of validity.

Solution

\(3\left(1 - \frac{x}{9}\right)^{\frac{1}{2}} = 3\left(1 - \dfrac{x}{18} - \dfrac{x^2}{648}\right) = 3 - \dfrac{x}{6} - \dfrac{x^2}{216} + \cdots\), \(\;|x| < 9\). See 4.3.

10 Paper 1 AHL (a) Express \(\dfrac{1}{(1 - x)(1 + 2x)}\) in partial fractions.

(b) Hence expand it up to the term in \(x^2\), and state the interval of validity.

Solution

(a) \(1 \equiv A(1 + 2x) + B(1 - x)\): \(x = 1\): \(A = \frac{1}{3}\); \(x = -\frac{1}{2}\): \(B = \frac{2}{3}\). \(\;\dfrac{1}{3(1 - x)} + \dfrac{2}{3(1 + 2x)}\).

(b) \(\dfrac{1}{3}(1 + x + x^2) + \dfrac{2}{3}(1 - 2x + 4x^2) = 1 - x + 3x^2 + \cdots\), valid for \(|x| < \dfrac{1}{2}\). See 4.3 and 3.1.