Use the following formulas: \begin{align} n! & = n \times (n - 1) \times ... \times 2 \times 1 \\ \\ \binom{n}{r}& ={n!\over r!(n-r)!}={n(n-1) \ldots (n-r+1)\over r!} \end{align}

(a)

\begin{align} 2! \times 3! & = (2 \times 1) \times (3 \times 2 \times 1) \\ & = 2 \times 6 \\ & = 12 \end{align}

(b)

\begin{align} \require{cancel} {6! \over 3!} & = {6 \times 5 \times 4 \times \cancel{3 \times 2 \times 1} \over \cancel{3 \times 2 \times 1}} \\ & = 6 \times 5 \times 4 \\ & = 120 \end{align}

(c)

\begin{align} \require{cancel} \binom{5}{1} & = {5! \over 1!(5 - 1)!} \\ & = {5! \over 1! \phantom{0} 4!} \\ & = {5 \times \cancel{4 \times 3 \times 2 \times 1} \over (1)(\cancel{4 \times 3 \times 2 \times 1})} \\ & = {5 \over 1} \\ & = 5 \end{align}

(d)

\begin{align} \require{cancel} \binom{9}{2} & = {9! \over 2!(9 - 2)!} \\ & = {9 \times 8 \times \cancel{7 \times 6 \times ... \times 2 \times 1} \over (2 \times 1)(\cancel{7 \times 6 \times ... \times 2 \times 1})} \\ & = {9 \times 8 \over 2} \\ & = 36 \end{align}

Pascal’s Triangle: \begin{align} & 1 & (n = 0)\\ 1 & \phantom{0} 1 & (n = 1) \\ 1 \phantom{0} & 2 \phantom{0} 1 & (n = 2) \\ 1 \phantom{0} 3 & \phantom{0} 3 \phantom{0} 1 & (n = 3) \\ 1 \phantom{0} 4 \phantom{0} & 6 \phantom{0} 4 \phantom{0} 1 & (n = 4) \\ 1 \phantom{0} 5 \phantom{0} 10 & \phantom{0} 10 \phantom{0} 5 \phantom{0} 1 & (n = 5) \end{align}

(a) Since $ n = 4 $, the coefficients are $ 1 \phantom{0} 4 \phantom{0} 6 \phantom{0} 4 \phantom{0} 1 $

\begin{align} (1 - 2x)^4 & = 1(-2x)^0 + 4(-2x)^1 + 6(-2x)^2 + 4(-2x)^3 + 1(-2x)^4 \\ & = 1(1) + 4(-2x) + 6(4x^2) + 4(-8x^3) + 1(16x^4) \\ & = 1 - 8x + 24x^2 - 32x^3 + 16x^4 \end{align}

(b) Since $ n = 5 $, the coefficients are $ 1 \phantom{0} 5 \phantom{0} 10 \phantom{0} 10 \phantom{0} 5 \phantom{0} 1$

\begin{align} (1 + 3x)^5 & = 1(3x)^0 + 5(3x)^1 + 10(3x)^2 + 10(3x)^3 + 5(3x)^4 + 1(3x)^4 \\ & = 1(1) + 5(3x) + 10(9x^2) + 10(27x^3) + 5(81x^4) + 1(243x^5) \\ & = 1 + 15x + 90x^2 + 270x^3 + 405x^4 + 243x^5 \end{align}

Pascal’s Triangle: \begin{align} & 1 & (n = 0)\\ 1 & \phantom{0} 1 & (n = 1) \\ 1 \phantom{0} & 2 \phantom{0} 1 & (n = 2) \\ 1 \phantom{0} 3 & \phantom{0} 3 \phantom{0} 1 & (n = 3) \\ 1 \phantom{0} 4 \phantom{0} & 6 \phantom{0} 4 \phantom{0} 1 & (n = 4) \\ 1 \phantom{0} 5 \phantom{0} 10 & \phantom{0} 10 \phantom{0} 5 \phantom{0} 1 & (n = 5) \\ 1 \phantom{0} 6 \phantom{0} 15 \phantom{0} & 20 \phantom{0} 15 \phantom{0} 6 \phantom{0} 1 & (n = 6) \\ 1 \phantom{0} 7 \phantom{0} 21 \phantom{0} 35 & \phantom{0} 35 \phantom{0} 21 \phantom{0} 7 \phantom{0} 1 & (n = 7) \\ 1 \phantom{0} 8 \phantom{0} 28 \phantom{0} 56 \phantom{0} & 70 \phantom{0} 56 \phantom{0} 28 \phantom{0} 8 \phantom{0} 1 & (n = 8) \\ 1 \phantom{0} 9 \phantom{0} 36 \phantom{0} 84 \phantom{0} 126 & \phantom{0} 126 \phantom{0} 84 \phantom{0} 36 \phantom{0} 9 \phantom{0} 1 & (n = 9) \\ 1 \phantom{0} 10 \phantom{0} 45 \phantom{0} 120 \phantom{0} 210 \phantom{0} & 252 \phantom{0} 210 \phantom{0} 120 \phantom{0} 45 \phantom{0} 10 \phantom{0} 1 & (n = 10) \end{align}

(a)Since $n = 9$, the first four coefficients are $1 \phantom{0} 9 \phantom{0} 36 \phantom{0} 84$

\begin{align} (1 + 2x)^9 & = 1(2x)^0 + 9(2x)^1 + 36(2x)^2 + 84(2x)^3 + ... \\ & = 1(1) + 9(2x) + 36(4x^2) + 84(8x^3) + ... \\ & = 1 + 18x + 144x^2 + 672x^3 + ... \end{align}

(b) Since $n = 10$, the first four coefficients are $1 \phantom{0} 10 \phantom{0} 45 \phantom{0} 120$

\begin{align} (1 -x^2)^{10} & = 1(-x^2)^0 + 10(-x^2)^1 + 45(-x^2)^2 + 120(-x^2)^3 + ... \\ & = 1(1) + 10(-x^2) + 45(x^4) + 120(-x^6) + ... \\ & = 1 - 10x^2 + 45x^4 - 120x^6 + ... \end{align}

(a)

\begin{align} (1 + x)^{10} & = (1)^{10} + \binom{10}{1}(1)^9(x)^1 + \binom{10}{2}(1)^8(x)^2 + \binom{10}{3}(1)^7(x)^3 + ... \\ & = 1 + (10)(1)(x) + (45)(1)(x^2) + (120)(1)(x^3) + ... \\ & = 1 + 10x + 45x^2 + 120x^3 +... \end{align}

(b)

\begin{align} (1 - 3x)^8 & = [1 + (-3x)]^8 \\ & = (1)^8 + \binom{8}{1}(1)^7(-3x)^1 + \binom{8}{2}(1)^6(-3x)^2 + \binom{8}{3}(1)^5(-3x)^3 + ... \\ & = 1 + (8)(1)(-3x) + (28)(1)(9x^2) + (56)(1)(-27x^3) + ... \\ & = 1 - 24x + 252x^2 - 1512x^3 + ... \end{align}

(c)

\begin{align} (1 - 2x^2)^7 & = [1 + (-2x^2)]^7 \\ & = (1)^7 + \binom{7}{1}(1)^6(-2x^2)^1 + \binom{7}{2}(1)^5(-2x^2)^2 + \binom{7}{3}(1)^4(-2x^2)^3 + ... \\ & = 1 + (7)(1)(-2x^2) + (21)(1)(4x^4) + (35)(1)(-8x^6) + ... \\ & = 1 - 14x^2 + 84x^4 - 280x^6 + ... \end{align}

(d)

\begin{align} \left(1 - {1 \over 2}x^3\right)^{16} & = \left[ 1 + \left(-{1 \over 2}x^3\right) \right]^{16} \\ & = (1)^{16} + \binom{16}{1}(1)^{15}\left(-{1 \over 2}x^3\right)^1 + \binom{16}{2}(1)^{14}\left(-{1 \over 2}x^3\right)^2 + \binom{16}{3}(1)^{13}\left(-{1 \over 2}x^3\right)^3 + ... \\ & = 1 + (16)(1)\left(-{1 \over 2}x^3\right) + (120)(1)\left({1 \over 4}x^6\right) + (560)(1)\left(-{1 \over 8}x^9\right) + ... \\ & = 1 - 8x^3 + 30x^6 - 70x^9 + ... \end{align}

Use the formulas: \begin{align} n! & = n \times (n - 1) \times ... \times 2 \times 1 \\ \\ \binom{n}{r}& ={n!\over r!(n-r)!}={n(n-1) \ldots (n-r+1)\over r!} \end{align}

(i)

\begin{align} \require{cancel} \binom{n}{2} & = {n! \over 2! (n - 2)!} \\ & = {n \times (n - 1) \times \cancel{(n - 2) \times ... \times 2 \times 1} \over (2 \times 1)[\cancel{(n - 2) \times ... \times 2 \times 1}]} \\ & = {n(n - 1) \over 2} \\ \\ \\ \binom{n + 1}{3} & = {(n + 1)! \over 3!(n + 1 - 3)!} \\ & = {(n + 1)! \over 3!(n - 2)!} \\ & = {(n + 1) \times n \times (n - 1) \times \cancel{(n - 2) \times ... \times 2 \times 1} \over (3 \times 2 \times 1)[\cancel{(n - 2) \times ... \times 2 \times 1]}} \\ & = {(n + 1)(n)(n - 1) \over 6} \end{align}

(ii)

\begin{align} \require{cancel} \binom{n + 1}{3} & = 4\binom{n}{2} \\ \\ \text{Using } & \text{results from (i),} \\ \\ {(n + 1)n(n - 1) \over 6} & = 4 \left[ n(n - 1) \over 2 \right] \\ (n + 1)n(n - 1) & = 24 \left[ n(n - 1) \over 2\right] \\ (n + 1)\cancel{n}\cancel{(n - 1)} & = 12\cancel{n}\cancel{(n - 1)} \\ n + 1 & = 12 \\ n & = 12 - 1 \\ & = 11 \end{align}

(i)

\begin{align} (1 + \sqrt{x})^5 & = (1)^5 + \binom{5}{1} (1)^4 (\sqrt{x})^1 + \binom{5}{2} (1)^3 (\sqrt{x})^2 + \binom{5}{3} (1)^2 (\sqrt{x})^3 \\ & \phantom{=} + \binom{5}{4} (1)^1 (\sqrt{x})^4 + \binom{5}{5} (1)^0 (\sqrt{x})^5 \\ \\ & = 1 + (5)(1)(\sqrt{x}) + (10)(1)(x) + (10)(1)(x\sqrt{x}) + (5)(1)(x^2) + (1)(1)(x^2\sqrt{x}) \\ & = 1 + 5\sqrt{x} + 10x + 10x\sqrt{x} + 5x^2 + x^2\sqrt{x} \\ \\ (1 - \sqrt{x})^5 & = (1)^5 + \binom{5}{1} (1)^4 (-\sqrt{x})^1 + \binom{5}{2} (1)^3 (-\sqrt{x})^2 + \binom{5}{3} (1)^2 (-\sqrt{x})^3 \\ & \phantom{=} + \binom{5}{4} (1)^1 (-\sqrt{x})^4 + \binom{5}{5} (1)^0 (-\sqrt{x})^5 \\ \\ & = 1 + (5)(1)(-\sqrt{x}) + (10)(1)(x) + (10)(1)(-x\sqrt{x}) + (5)(1)(x^2) + (1)(1)(-x^2\sqrt{x}) \\ & = 1 - 5\sqrt{x} + 10x - 10x\sqrt{x} + 5x^2 - x^2\sqrt{x} \\ \\ \\ \therefore & \phantom{00} (1 + \sqrt{x})^5 - (1 - \sqrt{x})^5 \\ \\ & = (1 + 5\sqrt{x} + 10x + 10x\sqrt{x} + 5x^2 + x^2\sqrt{x}) - (1 - 5\sqrt{x} + 10x - 10x\sqrt{x} + 5x^2 - x^2\sqrt{x}) \\ \\ & = 1 + 5\sqrt{x} + 10x + 10x\sqrt{x} + 5x^2 + x^2\sqrt{x} - 1 + 5\sqrt{x} - 10x + 10x\sqrt{x} - 5x^2 + x^2\sqrt{x} \\ \\ & = 1 - 1 + 5\sqrt{x} + 5\sqrt{x} + 10x - 10x + 10x\sqrt{x} + 10x\sqrt{x} + 5x^2 - 5x^2 + x^2\sqrt{x} + x^2\sqrt{x} \\ \\ & = 10\sqrt{x} + 20x\sqrt{x} + 2x^2\sqrt{x} \end{align}

(ii)

\begin{align} \text{From } & \text{part (i),} \\ (1 + \sqrt{x})^5 - (1 - \sqrt{x})^5 & = 10\sqrt{x} + 20x\sqrt{x} + 2x^2\sqrt{x} \\ \\ \text{When } & x = 2, \\ (1 + \sqrt{2})^5 - (1 - \sqrt{2})^5 & = 10\sqrt{2} + 20(2)\sqrt{2} + 2(2)^2\sqrt{2} \\ & = 10\sqrt{2} + 40\sqrt{2} + 8\sqrt{2} \\ & = 58\sqrt{2} \end{align}

(i) Take note only the terms up to x³ are required

\begin{align} \left(1 + {1 \over 2}x\right)^8 & = (1)^8 + \binom{8}{1} (1)^7 \left({1 \over 2}x\right)^1 + \binom{8}{2} (1)^6 \left({1 \over 2}x\right)^2 + \binom{8}{3} (1)^5 \left({1 \over 2}x\right)^3 + ... \\ & = 1 + (8)(1)\left({1 \over 2}x\right) + (28)(1)\left({1 \over 4}x^2\right) + (56)(1)\left({1 \over 8}x^3\right) + ... \\ & = 1 + 4x + 7x^2 + 7x^3 + ... \end{align} \begin{align} (2 - x)\left(1 + {1 \over 2}x\right)^8 & = (2 - x)(1 + 4x + 7x^2 + 7x^3 + ...) \\ & = 2 + 8x + 14x^2 + 14x^3 - x - 4x^2 - 7x^3 + ... \\ & = 2 + 7x + 10x^2 + 7x^3 + ... \end{align}

(ii)

\begin{align} (2 - x)\left(1 + {1 \over 2}x\right)^8 & = 1. 9 \times (1.05)^8 \\ \\ 2 - x & = 1.9 \\ 2 - 1.9 & = x \\ 0.1 & = x \\ \\ \text{From } & \text{part (i),} \\ (2 - x)\left(1 + {1 \over 2}x\right)^8 & = 2 + 7x + 10x^2 + 7x^3 + ... \\ \\ \text{When } & x = 0.1, \\ 1.9 \times (1.05)^8 & = 2 + 7(0.1) + 10(0.1)^2 + 7(0.1)^3 + ... \\ & \approx 2.807 \end{align}

(i)

\begin{align} (1 + x)^{10 \phantom{.} 000} & = (1)^{10 \phantom{.} 000} + \binom{10 \phantom{.} 000}{1}(1)^{9 \phantom{.} 999}(x)^{1} + ... \\ & = 1 + (10 \phantom{.} 000)(1)(x) + ... \\ & = 1 + 10 \phantom{.} 000x + ... \end{align}

(ii)

\begin{align} (1 + x)^{10 \phantom{.} 000} & = 1.1^{10 \phantom{.} 000} \\ \\ 1 + x & = 1.1 \\ x & = 1.1 - 1 \\ x & = 0.1 \\ \\ \text{From } & \text{part (i),} \\ (1 + x)^{10 \phantom{.} 000} & = 1 + 10 \phantom{.} 000x + ... \\ \\ \text{When } & x = 0.1, \\ 1.1^{10 \phantom{.} 000} & = 1 + 10 \phantom{.} 000(0.1) + ... \\ & \approx 1001 \\ \\ \\ \therefore 1.1^{10 \phantom{.} 000} & \text{ is larger} \end{align}

(i)

\begin{align} (1 + x)^{20} & = (1)^{20} + \binom{20}{1}(1)^{19}(x)^1 + \binom{20}{2}(1)^{18}(x)^2 + ... \\ & = 1 + (20)(1)(x) + (190)(1)(x^2) + ... \\ & = 1 + 20x + 190x^2 + ... \\ \\ 20 \phantom{.} 000\left(1 + {r \over 400}\right)^{4t} & = 20 \phantom{.} 000 (1 + x)^{20} \\ 20 \phantom{.} 000\left(1 + {4 \over 400}\right)^{4(5)} & = 20 \phantom{.} 000(1 + x)^{20} \\ 20 \phantom{.} 000\left(1 + 0.01\right)^{20} & = 20 \phantom{.} 000(1 + x)^{20} \\ \\ x & = 0.01 \\ \\ \text{Since } (1 + x)^{20} & = 1 + 20x + 190x^2 + ..., \\ 20 \phantom{.} 000(1 + x)^{20} & = 20 \phantom{.} 000(1 + 20x + 190x^2 +...) \\ \\ \text{When } & x = 0.01, \\ 20 \phantom{.} 000(1 + 0.01)^{20} & = 20 \phantom{.} 000[1 + 20(0.01) + 190(0.01)^2 + ...] \\ & \approx 24 \phantom{.} 380 \\ \\ \therefore \text{Amount of money received} & = \$ 24 \phantom{.} 380 \end{align}

(ii) Calculate the actual amount $A$, using the formula provided at the start of the question

\begin{align} \text{When } r = 4 & \text{ and } t = 5, \\ A & = 20 \phantom{.} 000 \left[1 + {(4) \over 400} \right]^{4(5)} \\ & = 24 \phantom{.} 403.80 \\ \\ {A - E \over A} \times 100\% & = {24 \phantom{.} 403.80 - 24 \phantom{.} 380 \over 24 \phantom{.} 403.80} \times 100\% \\ & = 0.097 \phantom{.} 525 \\ & \approx 0.10 \% \text{ (2 d.p.)} \end{align}

(iii)

Yes, the estimate in part (i) is a good estimate (as the percentage error is very low).

(i)(a)

\begin{align} (1 + 4x)^6 & = (1)^6 + \binom{6}{1}(1)^5(4x)^1 + \binom{6}{2}(1)^4(4x)^2 + ... \\ & = 1 + (6)(1)(4x) + (15)(1)(16x^2) + ... \\ & = 1 + 24x + 240x^2 + ... \end{align}

(i)(b)

\begin{align} (1 - 2x)^{14} & = (1)^{14} + \binom{14}{1}(1)^{13}(-2x)^1 + \binom{14}{2}(1)^{12}(-2x)^2 + ... \\ & = 1 + (14)(1)(-2x) + (91)(1)(4x^2) + ... \\ & = 1 - 28x + 364x^2 + ... \end{align}

(ii) Note only the first three terms are required

\begin{align} (1 + 4x)^6 (1 - 2x)^{14} & = (1 + 24x + 240x^2 + ...)(1 - 28x + 364x^2 + ...) \\ & = (1)(1) + (1)(-28x) + (1)(364x^2) + (24x)(1) + (24x)(-28x) + ... + (240x^2)(1) + ... \\ & = 1 - 28x + 364x^2 + 24x - 672x^2 + 240x^2 + ... \\ & = 1 - 28x + 24x + 364x^2 - 672x^2 + 240x^2 + ... \\ & = 1 - 4x - 68x^2 + ... \end{align}

(i)

\begin{align} (1 - x)^6 & = (1)^6 + \binom{6}{1} (1)^5 (-x)^1 + \binom{6}{2} (1)^4 (-x)^2 + \binom{6}{3} (1)^3 (-x)^3 + ... \\ & = 1 + (6)(1)(-x) + (15)(1)(x^2) + (20)(1)(-x^3) + ... \\ & = 1 - 6x + 15x^2 - 20x^3 + ... \end{align}

(ii)

\begin{align} (1 + x - 2x^2)^6 & = [1 - (-x + 2x^2)]^6 \\ & = [1 - (2x^2 - x)]^6 \end{align} \begin{align} \text{From } & \text{part (i),} \\ (1 - x)^6 & = 1 - 6x + 15x^2 - 20x^3 + ... \\ \\ \text{Let } & x = 2x^2 - x, \\ (1 + x - 2x^2)^6 & = 1 - 6(2x^2 - x) + 15(2x^2 - x)^2 - 20(2x^2 - x)^3 + ... \\ & = 1 - 6(2x^2 - x) + 15(4x^4 - 4x^3 + x^2) - 20(4x^4 - 4x^3 + x^2)(2x^2 - x) + ... \\ & = 1 - 6(2x^2 - x) + 15(4x^4 - 4x^3 + x^2) - 20(... - x^3) \phantom{00000} [\text{Ignore powers higher than } x^3] \\ & = 1 - 12x^2 + 6x + 60x^4 - 60x^3 + 15x^2 + 20x^3 + ... \\ & = 1 + 6x - 12x^2 + 15x^2 - 60x^3 + 20x^3 + ... \phantom{000000000000} [\text{Ignore powers higher than } x^3] \\ & = 1 + 6x + 3x^2 - 40x^3 + ... \end{align}

\begin{align} (1 + ax)^6 & = (1)^6 + \binom{6}{1} (1)^5 (ax)^1 + \binom{6}{2} (1)^4 (ax)^2 + ... \\ & = 1 + (6)(1)(ax) + (15)(1)(a^2x^2) + ... \\ & = 1 + 6ax + 15a^2 x^2 + ... \\ \\ (1 - x)(1 + ax)^6 & = (1 - x)(1 + 6ax + 15a^2x^2 + ...) \\ & = (1)(1) + (1)(6ax) + (1)(15a^2x^2) + (-x)(1) + (-x)(6ax) + ... \phantom{00000} [\text{Ignore terms higher than } x^2] \\ & = 1 + 6ax + 15a^2x^2 - x - 6ax^2 + ... \\ & = 1 + 6ax - x + 15a^2x^2 - 6ax^2 + ... \\ & = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\ \\ \therefore 1 + bx^2 & = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\ 1 + 0x + bx^2 & = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\ \\ \text{Comparing } & \text{coefficient of } x, \\ 0 & = 6a - 1 \\ 1 & = 6a \\ {1 \over 6} & = a \\ \\ \text{Comparing } & \text{coefficient of } x^2, \\ b & = 15a^2 - 6a \\ b & = 15 \left(1 \over 6\right)^2 - 6\left(1 \over 6\right) \phantom{00000000} \left[ \text{Substitute } a = {1 \over 6} \right] \\ b & = -{7 \over 12} \\ \\ \therefore a & = {1 \over 6}, b = -{7 \over 12} \end{align}

(i)

\begin{align} (1 - 2x)^n & = (1)^n + \binom{n}{1}(1)^{n - 1} (-2x)^1 + \binom{n}{2} (1)^{n - 2} (-2x)^2 + ... \\ & = 1 + \binom{n}{1}(1)(-2x) + \binom{n}{2} (1)(4x^2) + ... \\ & = 1 - 2\binom{n}{1}x + 4 \binom{n}{2}x^2 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = 4 \binom{n}{2} \end{align}

(ii)

\begin{align} 4\binom{n}{2} & = 24 \\ \binom{n}{2} & = 6 \\ {n(n - 1) \over 2!} & = 6 \\ {n(n - 1) \over 2} & = 6 \\ n(n - 1) & = 2(6) \\ n(n - 1) & = 12 \\ n^2 - n & = 12 \\ n^2 - n - 12 & = 0 \\ n^2 - n - 12 & = 0 \\ (n - 4)(n + 3) & = 0 \\ \\ n - 4 = 0 \phantom{00} &\text{or}\phantom{0000} n + 3 = 0 \\ n = 4 \phantom{00} & \phantom{or0000+3} n = -3 \text{ (Reject, since } n > 0) \end{align}

(i)

\begin{align} (1 - x)^4 & = (1)^4 + {4 \choose 1} (1)^3 (-x) + {4 \choose 2}(1)^2 (-x)^2 + {4 \choose 3}(1)^1 (-x)^3 + (-x)^4 \\ & = 1 + (4)(1)(-x) + (6)(1)(x^2) + (4)(1)(-x^3) + x^4 \\ & = 1 - 4x + 6x^2 - 4x^3 + x^4 \end{align}

(ii)

\begin{align} \text{From (i), } & 1 - 4x + 6x^2 - 4x^3 + x^4 = (1 - x)^4 \\ \\ S & = 1 - 4(1 - x^3) + 6(1 - x^3)^2 - 4(1 - x^3)^3 + (1 - x^3)^4 \\ & = [1 - (1 - x^3)]^4 \\ & = (1 - 1 + x^3)^4 \\ & = (x^3)^4 \\ & = x^{12} \end{align}

(i)

\begin{align} (1 + x)^5 & = (1)^5 + {5 \choose 1} (1)^4 (x) + {5 \choose 2}(1)^3 (x)^2 + ... \\ & = 1 + (5)(1)(x) + (10)(1)(x^2) + ... \\ & = 1 + 5x + 10x^2 + ... \\ \\ (1 - 4x)^4 & = (1)^4 + {4 \choose 1}(1)^3 (-4x) + {4 \choose 2} (1)^2 (-4x)^2 + ... \\ & = 1 + (4)(1)(-4x) + (6)(1)(16x^2) + ... \\ & = 1 - 16x + 96x^2 + ... \\ \\ (1 + x)^5 (1 - 4x)^4 & = (1 + 5x + 10x^2 + ...)(1 - 16x + 96x^2 + ...) \\ & = (1)(1) + (1)(-16x) + (1)(96x^2) + (5x)(1) + (5x)(-16x) + ... + (10x^2)(1) + ... \\ & = 1 - 16x + 96x^2 + 5x - 80x^2 + 10x^2 + ... \\ & = 1 - 11x + 26x^2 + ... \\ & \approx 1 - 11x + 26x^2 \end{align}

(ii)(a)

\begin{align} (1 + x)^5 (1 - 4x)^5 & = [(1 + x)^5 (1 - 4x)^4] (1 - 4x) \\ & \approx (1 - 11x + 26x^2) (1 - 4x) \\ & = (1)(1) + (1)(-4x) + (-11x)(1) + (-11x)(-4x) + (26x^2)(1) + ... \\ & = 1 - 4x - 11x + 44x^2 + 26x^2 + ... \\ & = 1 - 15x + 70x^2 +... \end{align}

(ii)(b)

\begin{align} \text{From (i), } (1 + x)^5 (1 - 4x)^4 & \approx 1 - 11x + 26x^2 \\ \\ \text{Replace } & x \text{ by } - x, \\ [1 + (-x)]^5 [1 - 4(-x)]^4 & \approx 1 - 11(-x) + 26(-x)^2 \\ (1 - x)^5 (1 + 4x)^4 & \approx 1 + 11x + 26x^2 \end{align}

(ii)(c)

\begin{align} (1 + x^2)^5 (1 - 2x)^4 (1 + 2x)^4 & = (1 + x^2)^5 [(1 - 2x)(1 + 2x)]^4 \phantom{000000} [(a - b)(a + b) = a^2 - b^2] \\ & = (1 + x^2)^5 [(1)^2 - (2x)^2]^4 \\ & = (1 + x^2)^5 (1 - 4x^2)^4 \\ \\ \text{From (i), } (1 + x)^5 (1 - 4x)^4 & \approx 1 - 11x + 26x^2 \\ \\ \text{Replace } & x \text{ by } x^2, \\ (1 + x^2)^5 (1 - 4x^2)^4 & \approx 1 - 11(x^2) + 26(x^2)^2 \\ & = 1 - 11x^2 + 26x^4 \end{align}

\begin{align} 28x^6 - & 30x^4 - 1 = 0 \\ \\ \text{Let } & x = 1 + h, \\ 28(1 + h)^6 - & 30(1 + h)^4 - 1 = 0 \\ \\ (1 + h)^6 & = (1)^6 + {6 \choose 1}(1)^5 (h) + {6 \choose 2}(1)^4 (h)^2 + ... \\ & = 1 + (6)(1)(h) + (16)(1)(h^2) + ... \\ & = 1 + 6h + 16h^2 + ... \\ \\ (1 + h)^4 & = (1)^4 + {4 \choose 1}(1)^3 (h) + {4 \choose 2}(1)^2 (h)^2 + ... \\ & = 1 + (4)(1)(h) + (6)(1)(h^2) + ... \\ & = 1 + 4h + 6h^2 + ... \end{align}
\begin{align} 28(1 + h)^6 - 30(1 + h)^4 - 1 & = 0 \\ 28(1 + 6h + 16h^2) - 30(1 + 4h + 6h^2) - 1 & = 0 \\ 28 + 168h + 448h^2 - 30 - 120h - 180h^2 - 1 & = 0 \\ 268h^2 + 48h - 3 & = 0 \\ \\ h & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(48) \pm \sqrt{48^2 - 4(268)(-3)} \over 2(268)} \\ & = {- 48 \pm \sqrt{5520} \over 536} \\ & = 0.049 \phantom{.} 061 \text{ or } -0.228 \phantom{.} 16 \\ \\ \text{If } & h = 0.049 \phantom{.} 061, \\ x & = 1 + 0.049 \phantom{.} 061 \\ x & = 1.049 \phantom{.} 061 \\ \\ \text{If } & h = -0.228 \phantom{.} 16, \\ x & = 1 + (-0.228 \phantom{.} 16) \\ x & = 0.771 \phantom{.} 84 \text{ (Reject, since } x > 1) \\ \\ \therefore x & \approx 1.05 \end{align}

(i)

\begin{align} (1 + 2x)^{2n} & = {2n \choose 0} (1)^{2n} (2x)^0 + {2n \choose 1} (1)^{2n - 1} (2x)^1 + {2n \choose 2} (1)^{2n - 2} (2x)^2 + ... + {2n \choose 2n} (1)^0 (2x)^{2n} \\ & = {2n \choose 0}(1)(1) + {2n \choose 1} (1)(2x) + {2n \choose 2} (1)(2x)^2 + ... + {2n \choose 2n} (1)(2x)^{2n} \\ & = {2n \choose 0} + {2n \choose 1}(2x) + {2n \choose 2}(2x)^2 + ... + {2n \choose 2n}(2x)^{2n} \end{align}

(ii)

\begin{align} (1 + 2x)^{2n} & = {2n \choose 0} + {2n \choose 1}(2x) + {2n \choose 2}(2x)^2 + ... + {2n \choose 2n}(2x)^{2n} \\ \\ \text{Let } & x = 0.5, \\ [1 + 2(0.5)]^{2n} & = {2n \choose 0} + {2n \choose 1}[2(0.5)] + {2n \choose 2}[2(0.5)]^2 + ... + {2n \choose 2n}[2(0.5)]^{2n} \\ 2^{2n} & = {2n \choose 0} + {2n \choose 1}(1) + {2n \choose 2}(1)^2 + ... + {2n \choose 2n}(1)^{2n} \\ & = {2n \choose 0} + {2n \choose 1} + {2n \choose 2} + ... + {2n \choose 2n} \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} (1 + y)^7 & = (1)^7 + {7 \choose 1}(1)^6(y) + {7 \choose 2}(1)^5 (y)^2 + {7 \choose 3} (1)^4 (y)^3 \\ & \phantom{00} + {7 \choose 4}(1)^3 (y)^4 + {7 \choose 5} (1)^2 (y)^5 + {7 \choose 6}(1)(y)^6 + (y)^7 \\ & = 1 + 7y + 21y^2 + 35y^3 + 35y^4 + 21y^5 + 7y^6 + y^7 \end{align}

(ii)

\begin{align} (1 + y)^7 & = 1 + 7y + 21y^2 + 35y^3 + 35y^4 + 21y^5 + 7y^6 + y^7 \\ \\ \text{Replace } & y \text{ by } x(1 + x), \\ [1 + x(1 + x)]^7 & = ... + 35[x(1 + x)]^4 + 21[x(1 + x)]^5 + 7[x(1 + x)]^6 + [x(1 + x)]^7 \\ (1 + x + x^2)^7 & = ... + 35x^4 (1 + x)^4 + 21x^5(1 + x)^5 + 7x^6(1 + x)^6 + x^7 (1 + x)^7 \\ \\ \\ \text{For } (1 + x)^4, T_{3 + 1} & = {4 \choose 3} (1)^{4 - 3} (x)^3 \\ & = (4)(1)(x^3) \\ & = 4x^3 \\ \\ 35x^4 (1 + x)^4 & = 35x^4 (... + 4x^3 + ...) \\ & = 140x^7 + ... \\ \\ \\ \text{For } (1 + x)^5, T_{2 + 1} & = {5 \choose 2} (1)^{5 - 2} (x)^2 \\ & = (10)(1)(x^2) \\ & = 10x^2 \\ \\ 21x^5(1 + x)^5 & = 21x^5 (... + 10x^2 + ...) \\ & = 210x^7 + ... \\ \\ \\ \text{For } (1 + x)^6, T_{1 + 1} & = {6 \choose 1} (1)^{6 - 1} (x) \\ & = (6)(1)(x) \\ & = 6x \\ \\ 7x^6 (1 + x)^6 & = 7x^6 (... + 6x + ...) \\ & = 42x^7 + ... \\ \\ \\ \text{Coefficient of } x^7 & = 140 + 210 + 42 + 1 \\ & = 393 \end{align}

(iii)

\begin{align} \text{From (i), } (1 + y)^7 & = 1 + 7y + 21y^2 + 35y^3 + 35y^4 + 21y^5 + 7y^6 + y^7 \\ \\ \text{Only first} & \text{ and last term that are not divisible by 7} \\ \\ \text{From (ii), } & \text{replace } y \text{ by } x(1 + x), \\ (1 + x + x^2)^7 & = ... + [x(1 + x)]^7 \\ & = ... + x^7 (1 + x)^7 \\ & = ... + x^7 (\underbrace{1}_\text{Not divisible} + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + \underbrace{x^7}_\text{Not divisible}) \\ \\ \therefore \text{There will be } & \text{two terms that are not divisible by 7} \end{align}