(i)(a)

\begin{align} (1 + 3x)^6 & = (1)^6 + \binom{6}{1}(1)^5(3x)^1 + \binom{6}{2}(1)^4(3x)^2 + \binom{6}{3}(1)^3(3x)^3 + ... \\ & = 1 + (6)(1)(3x) + (15)(1)(9x^2) + (20)(1)(27x^3) + ... \\ & = 1 + 18x + 135x^2 + 540x^3 + ... \end{align}

(i)(b)

\begin{align} (1 - 4x)^5 & = (1)^5 + \binom{5}{1}(1)^4(-4x)^1 + \binom{5}{2}(1)^3(-4x)^2 + \binom{5}{3}(1)^2(-4x)^3 + ... \\ & = 1 + (5)(1)(-4x) + (10)(1)(16x^2) + (10)(1)(-64x^3) + ... \\ & = 1 - 20x + 160x^2 - 640x^3 + ... \end{align}

(ii)

\begin{align} (1 + 3x)^6 (1 - 4x)^5 & = (1 + 18x + 135x^2 + 540x^3 + ...)(1 - 20x + 160x^2 - 640x^3 + ...) \\ & = ... + (1)(160x^2) + … + (18x)(-20x) + … + (135x^2)(1) + ... \\ & = 160x^2 - 360x^2 + 135x^2 + ... \\ & = -65x^2 + … \\ \\ \therefore \text{Coefficient of } x^2 & = -65 \end{align}

(i)

\begin{align} \text{General term, } T_{r + 1} & = \binom{n}{r} a^{n - r} b^r \\ & = \binom{8}{r}(1)^{8 - r} \left({x \over 2}\right)^r \\ & = \binom{8}{r}(1) \left({1 \over 2} \times x \right)^r \\ & = \binom{8}{r}(1)\left(1 \over 2\right)^r (x)^r \\ & = \binom{8}{r}\left(1 \over 2 \right)^r x^r \\ \\ \text{Let } r & = 3 \\ \\ T_{3 + 1} & = \binom{8}{3}\left(1 \over 2\right)^3 x^3 \\ & = (56)\left(1 \over 8\right) x^3 \\ & = 7x^3 \\ \\ \therefore \text{Coefficient of } x^3 & = 7 \end{align}

(ii)

\begin{align} \text{General term, } T_{r + 1} & = \binom{n}{r} a^{n - r} b^r \\ & = \binom{12}{r}(x)^{12 - r}\left(- {2 \over x^2}\right)^r \\ & = \binom{12}{r}(x^{12 - r}) \left( - 2 \times {1 \over x^2} \right)^r \\ & = \binom{12}{r} (x^{12 - r}) (-2)^r \left(1 \over x^2\right)^r \\ & = \binom{12}{r} (-2)^r (x^{12 - r})(x^{-2})^r \\ & = \binom{12}{r} (-2)^r (x^{12 - r})(x^{-2r}) \\ & = \binom{12}{r} (-2)^r x^{12 - r + (-2r)} \\ & = \binom{12}{r} (-2)^r x^{12 - 3r} \\ \\ \text{Let } 12 - 3r & = 3 \\ - 3r & = - 9 \\ r & = {-9 \over -3} \\ & = 3 \\ \\ T_{3 + 1} & = \binom{12}{3}(-2)^3 x^{12 - 3(3)} \\ & = (220)(-8)x^3 \\ & = -1760x^3 \\ \\ \therefore \text{Coefficient of } x^3 & = -1760 \end{align}

(i)

\begin{align} \left( x^2 - {2 \over x} \right)^8 & = (x^2)^8 + \binom{8}{1}(x^2)^7\left(- {2 \over x}\right)^1 + \binom{8}{2}(x^2)^6\left(-{2 \over x}\right)^2 + \binom{8}{3}(x^2)^5\left(-{2 \over x}\right)^3 + ... \\ & = x^{16} + (8)(x^{14})\left(-{2 \over x}\right) + (28)(x^{12})\left(4 \over x^2\right) + (56)(x^{10})\left(-{8 \over x^3}\right) + ... \\ & = x^{16} - (16)\left(x^{14} \over x\right) + (112)\left(x^{12} \over x^2\right) - (448)\left(x^{10} \over x^3\right) + ... \\ & = x^{16} - 16x^{13} + 112x^{10} - 448x^{7} + ... \end{align}

(ii)

\begin{align} (x^3 + 1)^2 \left(x^2 - {2 \over x}\right)^8 & = \left[ (x^3)^2 + 2(x^3)(1) + (1)^2 \right] \left(x^2 - {2 \over x}\right)^8 \\ & = (x^6 + 2x^3 + 1) (x^{16} - 16x^{13} + 112x^{10} - 448x^{7} + ...) \\ & = (x^6 + 2x^3 + 1)(x^{16} - 16x^{13} + 112x^{10} - 448x^{7} + ...) \\ & = ...+ (x^6)(-448x^7) + ... + (2x^3)(112x^{10}) + ... + (1)(-16x^{13}) + ... \\ & = -448x^{13} + 224x^{13} - 16x^{13} + ... \\ & = -240x^{13} + ... \\ \\ \therefore \text{Coefficient of } x^{13} & = -240 \end{align}

(i)

\begin{align} \left(1 - {x \over 2}\right)^n & = (1)^n + \binom{n}{1}(1)^{n - 1}\left(-{x \over 2}\right)^1 + \binom{n}{2}(1)^{n - 2}\left(-{x \over 2}\right)^2 + ... \\ & = 1 + (n)(1)\left(-{x \over 2}\right) + \left[{n(n - 1) \over 2}\right](1)\left(x^2 \over 4\right) + ... \\ & = 1 - \left({n \over 2}\right) x + \left[{n(n - 1) \over 8}\right] x^2 + ... \end{align}

(ii)

\begin{align} \text{Using the } & \text{result from part (i),} \\ 1 + ax + 7x^2 + ... & = 1 - \left({n \over 2}\right) x + \left[ {n(n - 1) \over 8}\right] x^2 + ... \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 7 & = {n (n - 1) \over 8} \\ 56 & = n(n - 1) \\ 56 & = n^2 - n \\ 0 & = n^2 - n - 56 \\ 0 & = (n - 8)(n + 7) \\ \\ n - 8 = 0 \phantom{000}&\text{or}\phantom{0000} n + 7 = 0 \\ n = 8 \phantom{000}& \phantom{or0000+7} n = - 7 \text{ (Reject)} \end{align}

(iii)

\begin{align} \text{From } & \text{part (ii),} \\ 1 + ax + 7x^2 + ... & = 1 - \left({n \over 2}\right) x + \left[ {n(n - 1) \over 8}\right] x^2 + ... \\ \\ \text{Comparing } & \text{coefficients of } x, \\ a & = -{n \over 2} \\ a & = -{8 \over 2} \phantom{00000000} [n = 8] \\ & = -4 \end{align}

(i)

\begin{align} (1 + p)^4 & = (1)^4 + \binom{4}{1}(1)^3(p)^1 + \binom{4}{2}(1)^2(p)^2 + \binom{4}{3}(1)^1(p)^3 + (p)^4 \\ & = 1 + (4)(1)(p) + (6)(1)(p^2) + (4)(1)(p^3) + p^4 \\ & = 1 + 4p + 6p^2 + 4p^3 + p^4 \end{align}

(ii)

\begin{align} \text{From } & \text{part (i),} \\ (1 + p)^4 & = 1 + 4p + 6p^2 + 4p^3 + p^4 \\ \\ \text{Let } & p = x + x^2, \\ (1 + x + x^2)^4 & = 1 + 4(x + x^2) + 6(x + x^2)^2 + 4(x + x^2)^3 + (x + x^2)^4 \\ & = 1 + 4x + 4x^2 + 6[x^2 + 2(x)(x^2) + (x^2)^2] + 4(x + x^2)^2(x + x^2) + ... \phantom{000000} [\text{Ignore powers higher than } x^3] \\ & = 1 + 4x + 4x^2 + 6(x^2 + 2x^3 + x^4) + 4(x^2 + 2x^3 + x^4)(x + x^2) + ... \\ & = 1 + 4x + 4x^2 + 6x^2 + 12x^3 + ... + 4(x^3 + ...) + ... \\ & = 1 + 4x + 4x^2 + 6x^2 + 12x^3 + 4x^3 + ... \\ & = 1 + 4x + 10x^2 + 16x^3 + ... \end{align}

(iii)

\begin{align} (1 + x + x^2)^4 & = (1.11)^4 \\ \\ 1 + x + x^2 & = 1.11 \\ x^2 + x & = 0.11 \\ 100x^2 + 100x & = 11 \phantom{00000} [\text{Multiply by 100 throughout}] \\ 100x^2 + 100x - 11 & = 0 \\ (10x - 1)(10x + 11) & = 0 \end{align} \begin{align} 10x - 1 & = 0 & 10x + 11 & = 0 \\ 10x & = 1 & 10x & = -11 \\ x & = {1 \over 10} & x & = -{11 \over 10} \\ & = 0.1 & & = -1.1 \end{align}
\begin{align} \text{From (ii), } (1 + x + x^2)^4 & = 1 + 4x + 10x^2 + 16x^3 + ... \\ \\ \text{For } & x = 0.1, \\ (1.11)^4 & = 1 + 4(0.1) + 10(0.1)^2 + 16(0.1)^3 + ... \\ & = 1.516 \\ \\ \text{For } & x = -1.1, \\ (1.11)^4 & = 1+ 4(-1.1) + 10(-1.1)^2 + 16(-1.1)^3 + ... \\ & = -12.596 \text{ (Reject, since } (1.11)^4 > 0) \\ \\ \therefore (1.11)^4 & \approx 1.516 \end{align}

(i)

\begin{align} (2 - x)^7 & = (2)^7 + \binom{7}{1} (2)^6 (-x)^1 + \binom{7}{2} (2)^5 (-x)^2 + ... \\ & = 128 + (7)(64)(-x) + (21)(32)(x^2) + ... \\ & = 128 - 448x + 672x^2 + ... \end{align}

(ii)

\begin{align} \text{Let } (2 - x)^7 & = 1.99^7 \\ \\ \therefore 2 - x & = 1.99 \\ - x & = -0.01 \\ x & = 0.01 \\ \\ \text{From } & \text{part (i),} \\ (2 - x)^7 & = 128 - 448x + 672x^2 + ... \\ \\ \text{Let } & x = 0.01, \\ (1.99)^7 & = 128 - 448(0.01) + 672(0.01)^2 + ... \\ & = 123.587 \end{align}

(iii)

\begin{align} (k - x)(2 - x)^7 & = (k - x)(128 - 448x + 672x^2 + ...) \\ & = ... + (k)(672x^2) + ... + (-x)(-448x) + ... \\ & = 672k x^2 + 448x^2 + ... \\ & = (672k + 448)x^2 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = 672k + 448 \\ 616 & = 672k + 448 \\ 616 - 448 & = 672k \\ 168 & = 627k \\ \\ k & = {168 \over 672} \\ & = {1 \over 4} \end{align}

(i)

\begin{align} \text{General term, } T_{r + 1} & = \binom{n}{r} a^{n - r} b^r \\ & = \binom{8}{r} \left(a^2 \over \sqrt{x}\right)^{8 - r} \left(- {\sqrt{x} \over a} \right)^r \\ & = \binom{8}{r} \left(a^2 \times {1 \over \sqrt{x}} \right)^{8 - r} \left( -{1 \over a} \times \sqrt{x} \right)^r \\ & = \binom{8}{r} (a^2)^{8 - r} \left(1 \over \sqrt{x}\right)^{8 - r} \left(-{1 \over a}\right)^r (\sqrt{x})^r \\ & = \binom{8}{r} (a^2)^{8 - r} \left(x^{-{1 \over 2}} \right)^{8 - r} \left(-{1 \over a}\right)^r \left(x^{1 \over 2}\right)^r \\ & = \binom{8}{r} (a^2)^{8 - r} \left(x^{-{1 \over 2}(8 - r)} \right) \left(-{1 \over a}\right)^r \left(x^{{1 \over 2}(r)}\right) \\ & = \binom{8}{r} (a^2)^{8 - r} \left(x^{-4 + {1 \over 2} r} \right) \left(-{1 \over a}\right)^r \left(x^{{1 \over 2}r}\right) \\ & = \binom{8}{r} (a^2)^{8 - r} \left(-{1 \over a}\right)^r \left(x^{-4 + {1 \over 2} r} \right) \left(x^{{1 \over 2}r}\right) \\ & = \binom{8}{r} (a^2)^{8 - r} \left(-{1 \over a}\right)^r \left(x^{-4 + {1 \over 2} r + {1 \over 2}r} \right) \\ & = \binom{8}{r} (a^2)^{8 - r} \left(-{1 \over a}\right)^r \left(x^{-4 + r}\right) \end{align}

There are a total of 9 terms in this expansion. Thus, the middle term is the 5th term
\begin{align} T_5 = T_{4 + 1} & = \binom{8}{4} (a^2)^{8 - 4} \left(-{1 \over a}\right)^4 \left(x^{-4 + 4}\right) \\ & = (70)(a^2)^4 \left(1 \over a^4\right)(x^0) \\ & = (70)(a^8)\left(1 \over a^4\right) (1) \\ & = (70)\left(a^8 \over a^4\right) \\ & = 70a^4 \end{align}

(ii) Note ${1 \over x} = x^{-1}$

\begin{align} \text{Let } -4 + r & = - 1 \\ r & = -1 + 4 \\ & = 3 \\ \\ T_{3 + 1} & = \binom{8}{3} (a^2)^{8 - 3} \left(-{1 \over a}\right)^3 \left(x^{-4 + 3} \right) \\ & = (56)(a^2)^5 \left(-{1 \over a^3}\right) (x^{-1}) \\ & = (56)(a^{10}) \left(-{1 \over a^3} \right) \left(1 \over x\right) \\ & = - 56 \left( {a^{10} \over a^3} \right) {1 \over x} \\ & = -56a^7 {1 \over x} \\ \\ \therefore \text{Coefficient of } {1 \over x} & = -56a^7 \end{align}

(i)

\begin{align} (2 + \sqrt{3})^5 & = (2)^5 + \binom{5}{1} (2)^4 (\sqrt{3})^1 + \binom{5}{2} (2)^3 (\sqrt{3})^2 + \binom{5}{3} (2)^2 (\sqrt{3})^3 + \binom{5}{4} (2)^1 (\sqrt{3})^4 + (\sqrt{3})^5 \\ & = 32 + (5)(16)(\sqrt{3}) + (10)(8)(3) + (10)(4)(3\sqrt{3}) + (5)(2)(9) + 9\sqrt{3} \\ & = 32 + 80\sqrt{3} + 240 + 120\sqrt{3} + 90 + 9\sqrt{3} \\ & = 32 + 240 + 90 + 80\sqrt{3} + 120\sqrt{3} + 9\sqrt{3} \\ & = 362 + 209\sqrt{3} \end{align}

(ii) The expansion is almost the same, except the 2nd, 4th and 6th term will be negative

\begin{align} (2 - \sqrt{3})^5 & = (2)^5 + \binom{5}{1} (2)^4 (-\sqrt{3})^1 + \binom{5}{2} (2)^3 (-\sqrt{3})^2 + \binom{5}{3} (2)^2 (-\sqrt{3})^3 + \binom{5}{4} (2)^1 (-\sqrt{3})^4 + (-\sqrt{3})^5 \\ & . \\ & . \\ & . \\ & = 32 - 80\sqrt{3} + 240 - 120\sqrt{3} + 90 - 9\sqrt{3} \\ & = 32 + 240 + 90 - 80\sqrt{3} - 120\sqrt{3} - 9\sqrt{3} \\ & = 362 - 209\sqrt{3} \end{align} \begin{align} {1 \over (2 + \sqrt{3})^5} & = {1 \over 362 + 209 \sqrt{3}} \phantom{00000} [\text{from part(i)}] \\ & = {1 \over 362 + 209 \sqrt{3}} \times {362 - 209\sqrt{3} \over 362 - 209\sqrt{3}} \phantom{00000} [\text{Rationalise denominator}] \\ & = {362 - 209\sqrt{3} \over (362 - 209\sqrt{3})(362 + 209\sqrt{3})} \\ & = {362 - 209\sqrt{3} \over (362)^2 - (209\sqrt{3})^2 } \\ & = {362 - 209\sqrt{3} \over 1} \\ & = 362 - 209 \sqrt{3} \\ & = (2 - \sqrt{3})^5 \text{ (Shown)} \end{align}

(i)

\begin{align} (a - x)(1 + 2x)^n & = 3 + 47x + bx^2 + ... \\ \\ \text{Let } & x = 0, \\ (a - 0)[1 + 2(0)]^n & = 3 + 47(0) + b(0)^2 + ... \\ (a)(1)^n & = 3 \\ (a)(1) & = 3 \\ a & = 3 \end{align}

(ii)

\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 + \binom{n}{1} 2x + \binom{n}{2} 4x^2 + ... \\ & = 1 + (n)(2x) + \left[{n(n - 1) \over 2}\right]4x^2 + ... \\ & = 1 + 2n x + 2n(n - 1)x^2 + ... \end{align} \begin{align} (3 - x)(1 + 2x)^n & = (3 - x)[1 + 2nx + 2n(n - 1)x^2 + ...] \\ & = (3)(1) + (3)(2nx) + (3)[2n(n - 1)x^2] + (-x)(1) + (-x)(2nx) + ... \\ & = 3 + 6nx + 6n(n - 1)x^2 - x - 2nx^2 + ... \\ & = 3 + 6nx + 6n^2x^2 - 6nx^2 - x - 2nx^2 + ... \\ & = 3 + 6nx - x + 6n^2x^2 - 6nx^2 - 2nx^2 + ... \\ & = 3 + (6n - 1)x + 6n^2 x^2 - 8nx^2 + ... \\ & = 3 + (6n - 1)x + (6n^2 - 8n) x^2 + ... \end{align} \begin{align} \therefore 3 + 47x + bx^2 & = 3 + (6n - 1)x + (6n^2 - 8n)x^2 \\ \\ \text{Comparing }& \text{coefficients of } x, \\ 47 & = 6n - 1 \\ 47 + 1 & = 6n \\ 48 & = 6n \\ \\ n & = {48 \over 6} \\ & = 8 \end{align}

(iii)

\begin{align} \text{From } & \text{part (ii),} \\ 3 + 47x + bx^2 & = 3 + (6n - 1)x + (6n^2 - 8n)x^2 \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ b & = 6n^2 - 8n \\ & = 6(8)^2 - 8(8) \phantom{00000000} [n = 8] \\ & = 320 \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 + ... \\ \\ (2 + bx)^5 & = (2)^5 + \binom{5}{1} (2)^4 (bx)^1 + \binom{5}{2} (2)^3 (bx)^2 + ... \\ & = 32 + (5)(16)(bx) + (10)(8)(b^2x^2) + ... \\ & = 32 + 80bx + 80b^2 x^2 + ... \\ \\ (1 + ax)^6(2 + bx)^5 & = (1 + 6ax + 15a^2 x^2 + ...)(32 + 80bx + 80b^2 x^2 + ...) \\ & = ... + (1)(80bx) + (1)(80b^2x^2) + (6ax)(32) + (6ax)(80bx) + ... + (15a^2x^2)(32) + ... \\ & = 80bx + 80b^2x^2 + 192ax + 480abx^2 + 480a^2 x^2 + ... \\ & = 80bx + 192ax + 80b^2 x^2 + 480abx^2 + 480a^2 x^2 + ... \\ & = (80b + 192a)x + (80b^2 + 480ab + 480a^2) x^2 + ... \end{align}
\begin{align} \text{Since the } & \text{coefficient of } x \text{is -112,} \\ -112 & = 80b + 192a \\ -7 & = 5b + 12a \\ \\ 5b & = -7 - 12a \\ \\ b & = {-7 - 12a \over 5} \text{ --- (1)} \\ \\ \text{Since the } & \text{coefficient of } x \text{is 80,} \\ 80 & = 80b^2 + 480ab + 480a^2 \\ 1 & = b^2 + 6ab + 6a^2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 1 & = \left(-7 - 12a \over 5\right)^2 + 6a\left(-7 - 12a \over 5\right) + 6a^2 \\ 1 & = {(- 7 - 12a)^2 \over 25} + {6a(- 7 - 12a) \over 5} + 6a^2 \\ 25 & = (-7-12a)^2 + 30a(-7-12a) + 150a^2 \\ 25 & = (-7)^2 + 2(-7)(-12a) + (-12a)^2 - 210a - 360a^2 + 150a^2 \\ 25 & = 49 + 168a + 144a^2 - 210a - 360a^2 + 150a^2 \\ 0 & = 49 - 25 + 168a - 210a + 144a^2 - 360a^2 + 150a^2 \\ 0 & = 24 - 42a - 66a^2 \\ 0 & = 66a^2 + 42a - 24 \\ 0 & = 11a^2 + 7a - 4 \\ 0 & = (a + 1)(11a - 4) \\ \\ a + 1 = 0 \phantom{000000.}&\text{or}\phantom{0000} 11a - 4 = 0 \\ a = -1 \phantom{00000}&\phantom{or0000+1} 11a = 4 \\ & \phantom{00000000000} a = {4 \over 11} \text{ (Reject, since } a \text{ is an integer)}\\ \\ \text{Substitute } & a = -1 \text{ into (1),} \\ b & = {-7 - 12(-1) \over 5} \\ & = 1 \\ \\ \therefore a & = -1, b = 1 \end{align}