(a)

\begin{align} (2 - x)^3 & = [2 + (-x)]^3 \\ & = (2)^3 + \binom{3}{1}(2)^2(-x)^1 + \binom{3}{2}(2)^1(-x)^2 + (-x)^3 \\ & = 8 + (3)(4)(-x) + (3)(2)(x^2) + (-x^3) \\ & = 8 - 12x + 6x^2 - x^3 \end{align}

(b)

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

(c)

\begin{align} (2 + x^2)^5 & = (2)^5 + \binom{5}{1}(2)^4(x^2)^1 + \binom{5}{2}(2)^3(x^2)^2 + \binom{5}{3}(2)^2(x^2)^3 + \binom{5}{4}(2)^1(x^2)^4 + (x^2)^5 \\ & = 32 + (5)(16)(x^2) + (10)(8)(x^4) + (10)(4)(x^6) + (5)(2)(x^8) + x^{10} \\ & = 32 + 80x^2 + 80x^4 + 40x^6 + 10x^8 + x^{10} \end{align}

(a)

\begin{align} \left(4 - {x \over 2}\right)^5 & = \left[ 4 - \left(x \over 2\right) \right]^5 \\ & = (4)^5 + \binom{5}{1}(4)^4\left(-{x \over 2}\right)^1 + \binom{5}{2}(4)^3\left(-{x \over 2}\right)^2 + \binom{5}{3}(4)^2\left(-{x \over 2}\right)^3 + ... \\ & = 1024 + (5)(256)\left(-{x \over 2}\right) + (10)(64)\left(x^2 \over 4\right) + (10)(16)\left(-{x^3 \over 8}\right) + ... \\ & = 1024 - 640x + 160x^2 - 20x^3 + ... \end{align}

(b)

\begin{align} \left({1 \over 2} + x^2\right)^{12} & = \left(1 \over 2\right)^{12} + \binom{12}{1}\left(1 \over 2\right)^{11}(x^2)^1 + \binom{12}{2}\left(1 \over 2\right)^{10}(x^2)^2 + \binom{12}{3}\left(1 \over 2\right)^9(x^2)^3 + ... \\ & = {1 \over 4096} + (12)\left(1 \over 2048\right)(x^2) + (66)\left(1 \over 1024\right)(x^4) + (220)\left(1 \over 512\right)(x^6) + ... \\ & = {1 \over 4096} + {3 \over 512}x^2 + {33 \over 512}x^4 + {55 \over 128}x^6 + ... \end{align}

(c)

\begin{align} \left({1 \over 2x} - 2x^2\right)^8 & = \left[ {1 \over 2x} + (-2x^2) \right]^8 \\ & = \left({1 \over 2x}\right)^8 + \binom{8}{1}\left(1 \over 2x\right)^7(-2x^2)^1 + \binom{8}{2}\left(1 \over 2x\right)^6(-2x^2)^2 + \binom{8}{3}\left(1 \over 2x\right)^5(-2x^2)^3 + ... \\ & = {1 \over 256x^8} + (8)\left(1 \over 128x^7\right)(-2x^2) + (28)\left(1 \over 64x^6\right)(4x^4) + (56)\left(1 \over 32x^5\right)(-8x^6) + ... \\ & = {1 \over 256x^8} - {1 \over 8x^5} + {7 \over 4x^2} - 14x + ... \\ & = {1 \over 256}x^{-8} - {1 \over 8}x^{-5} + {7 \over 4}x^{-2} - 14x + ... \end{align}

(a)

\begin{align} \text{General term } & = \binom{5}{r}(x)^{5 - r}(2y)^r \end{align}

(b)

\begin{align} \text{General term } & = \binom{10}{r}(2x)^{10 - r}(-3)^r \end{align}

(c)

\begin{align} \text{General term } & = \binom{13}{r}(3x^2)^{13 - r}\left(-{1 \over 2y} \right)^r \end{align}

(a)

\begin{align} \text{7th term (} r = 6 \text{), } \phantom{0} T_7 & = \binom{10}{6}(2)^4(x)^6 \\ & = (210)(16)(x^6) \\ & = 3 \phantom{.} 360x^6 \end{align}

(b)

\begin{align} \text{4th term (} r = 3 \text{), } \phantom{0} T_4 & = \binom{9}{3} (3x)^6 (-2)^3 \\ & = (84)(729x^6)(-8) \\ & = -489 \phantom{.} 888x^6 \end{align}

(c) Since there are 11 terms in total, the middle term is the 6th term

\begin{align} \text{6th term (} r = 5 \text{), } \phantom{0} T_6 & = \binom{10}{5} (y)^5 (-2x)^5 \\ & = (252) (y^5) (-32x^5) \\ & = -8 \phantom{.} 064y^5x^5 \end{align}

\begin{align} \text{General term } & = \binom{10}{r}(1)^{10 - r}\left( -{x^2 \over 4} \right)^r \\ & = \binom{10}{r} (1) \left( -{1 \over 4} \times x^2 \right)^r \\ & = \binom{10}{r}\left(-{1 \over 4}\right)^r (x^2)^r \\ & = \binom{10}{r}\left(-{1 \over 4}\right)^r (x^{2r}) \\ \\ \\ \text{Let } 2r & = 3 \\ r & = {3 \over 2} \\ & = 1.5 \\ \\ \text{Since } r \text{ is not an integer, } & \text{the term with } x^3 \text{ does not exist} \\ \\ \therefore \text{Coefficient of } x^3 & = 0 \\ \\ \\ \text{Let } 2r & = 6 \\ r & = 3 \\ \\ \text{4th term (} r = 3 \text{), } \phantom{0} T_4 & = \binom{10}{3}\left(-{1 \over 4}\right)^3 (x^{2(3)}) \\ & = (120)\left(-{1 \over 64}\right)(x^6) \\ & = -{15 \over 8}x^6 \\ \\ \text{Coefficient of } x^6 & = -{15 \over 8} \end{align}

(a) The term independent of x is the term with x⁰ or the constant term

\begin{align} \text{General term } & = \binom{18}{r} \left(1 \over x\right)^{18 - r} (-x^2)^r \\ & = \binom{18}{r} (x^{-1})^{18-r} (-1)^r (x^2)^r \\ & = \binom{18}{r} (x^{-(18-r)}) (-1)^r (x^{2r}) \\ & = \binom{18}{r} (x^{-18 + r}) (-1)^r (x^{2r}) \\ & = \binom{18}{r} (-1)^r (x^{-18+r + 2r}) \\ & = \binom{18}{r} (-1)^r (x^{-18+3r}) \\ \\ \text{Let } -18 + 3r & = 0 \\ 3r & = 18 \\ r & = 6 \\ \\ T_{6 + 1} = T_7 & = \binom{18}{6} (-1)^6 (x^{3(6) - 18}) \\ & = (18 \phantom{.} 564) (1) (x^0) \\ & = (18 \phantom{.} 564) (1) (1) \\ & = 18 \phantom{.} 564 \end{align}

(b) In this expansion, there is a total of 13 terms. So the middle term is the 7th term

\begin{align} T_{6 + 1} = T_7 & = \binom{12}{6} (x)^6 \left( 1 \over 2x^2 \right)^6 \\ & = (924)(x^6) \left[ 1^6 \over 2^6 (x^2)^6 \right] \\ & = (924)(x^6) \left(1 \over 64x^{12}\right) \phantom{000000} [ (a^m)^n = a^{mn}] \\ & = {924x^6 \over 64x^{12}} \\ & = {924 \over 64x^6} \\ & = {231 \over 16x^6} \end{align}

\begin{align} \require{cancel} (px - 3)^n & = (px)^n + \binom{n}{1} (px)^{n - 1}(-3)^1 + ... \\ & = p^n x^n + (n)(px)^{n - 1}(-3) + ... \\ & = p^n x^n + (n)(p^{n - 1}) (x^{n - 1}) (-3) + ... \\ & = p^n x^n + (n) (p^{n - 1}) (x^{n - 1}) (- 3) + ... \\ & = p^n x^n - 3np^{n - 1}x^{n - 1} + ... \\ \\ \therefore 512x^9 - qx^8 & = p^n x^n - 3np^{n-1}x^{n-1} \\ \\ \\ \text{Comparing } & \text{the first term,} \\ 512x^9 & = p^n x^n \\ \\ \therefore n & = 9 \\ \\ 512 & = p^n \phantom{00000000} [\text{Comparing coefficient}] \\ 512 & = p^9 \\ \sqrt[9]{512} & = p \\ 2 & = p \\ \\ \\ \text{Comparing } & \text{the second term,} \\ -qx^8 & = -3np^{n- 1}x^{n - 1} \\ -q\cancel{x^8} & = -3(9)(2)^8 \cancel{x^8} \phantom{0000000} [n = 9, p = 2]\\ -q & = -3(9)(2)^{8} \\ -q & = -6 \phantom{.} 912 \\ q & = 6 \phantom{.} 912 \\ \\ \therefore n & = 9, p = 2, q = 6 \phantom{.} 912 \end{align}

(i)

\begin{align} \left( 2 - {x \over 2} \right)^7 & = (2)^7 + \binom{7}{1} (2)^6 \left(-{x \over 2}\right)^1 + \binom{7}{2} (2)^5 \left(-{x \over 2}\right)^2 + \binom{7}{3} (2)^4 \left(-{x \over 2}\right)^3 + ... \\ & = 128 + (7)(64)\left(-{x \over 2}\right) + (21)(32)\left(x^2 \over 4\right) + (35)(16)\left(-{x^3 \over 8}\right) + ... \\ & = 128 - 224x + 168x^2 - 70x^3 + ... \end{align}

(ii)

\begin{align} \left( 2 - {x \over 2} \right)^7 & = (1.995)^7 \\ \\ \therefore 2 - {x \over 2} & = 1.995 \\ -{x \over 2} & = 1.995 - 2 \\ -{x \over 2} & = -0.005 \\ {x \over 2} & = 0.005 \\ x & = 2(0.005) \\ & = 0.01 \\ \\ \\ \text{From } & \text{part (i),} \\ \left( 2 - {x \over 2} \right)^7 & = 128 - 224x + 168x^2 - 70x^3 + ... \\ \\ \\ \text{Let } & x = 0.01, \\ (1.995)^7 & \approx 128 - 224(0.01) + 168(0.01)^2 - 70(0.01)^3 + ... \\ & = 125.776 \phantom{.} 73 \\ & \approx 125.776 \phantom{.} 7 \text{ (4 d.p.)} \end{align}

(i)

\begin{align} (1 - 2x)^9 & = (1)^9 + \binom{9}{1} (1)^8 (-2x)^1 + \binom{9}{2} (1)^7 (-2x)^2 + ... \\ & = 1 + (9)(1)(-2x) + (36)(1)(4x^2) + ... \\ & = 1 - 18x + 144x^2 + ... \\ \\ (2 + x)^5 & = (2)^5 + \binom{5}{1} (2)^4 (x)^1 + \binom{5}{2} (2)^3 (x)^2 + ... \\ & = 32 + (5)(16)(x) + (10)(8)(x^2) + ... \\ & = 32 + 80x + 80x^2 + ... \end{align}

(ii)

\begin{align} (1 - 2x)^9(2 + x)^5 & = (1 - 18x + 144x^2 + ...)(32 + 80x + 80x^2 + ...) \\ & = (1)(32) + (1)(80x) + (1)(80x^2) + (-18x)(32) + (-18x)(80x) + (144x^2)(32) + ... \\ & = 32 + 80x + 80x^2 - 576x - 1 \phantom{.} 440x^2 + 4 \phantom{.} 608x^2 + ... \\ & = 32 - 496x + 3 \phantom{.} 248x^2 + ... \end{align}

(i)

\begin{align} (2x - 1)^6 & = (2x)^6 + \binom{6}{1} (2x)^5 (-1)^1 + \binom{6}{2} (2x)^4 (-1)^2 + \binom{6}{3} (2x)^3 (-1)^3 + ... \\ & = 64x^6 + (6)(32x^5)(-1) + (15)(16x^4)(1) + (20)(8x^3)(-1) + ... \\ & = 64x^6 - 192x^5 + 240x^4 - 160x^3 + ... \end{align}

(ii)

\begin{align} (2x - 1)^6 (x^2 - 2x + 3) & = (64x^6 - 192x^5 + 240x^4 - 160x^3 + ...)(x^2 - 2x + 3) \\ & = ... + (-192x^5)(3) + ... + (240x^4)(-2x) + ... + (-160x^3)(x^2) + ... \\ & = -576x^5 - 480x^5 - 160x^5 + ... \\ & = -1216x^5 + ... \\ \\ \therefore \text{Coefficient of } x^5 & = -1216 \end{align}

(i)

\begin{align} \left( {1 \over 2} - 2x \right)^5 & = \left(1 \over 2\right)^5 + \binom{5}{1} \left(1 \over 2\right)^4 (-2x)^1 + \binom{5}{2} \left(1 \over 2\right)^3 (-2x)^2 + \binom{5}{3} \left(1 \over 2\right)^2 (-2x)^3 + ... \\ & = {1 \over 32} + (5)\left(1 \over 16\right)(-2x) + (10)\left(1 \over 8\right)(4x^2) + (10)\left(1 \over 4\right)(-8x^3) + ... \\ & = {1 \over 32} - {5 \over 8}x + 5x^2 - 20x^3 + ... \end{align}

(ii)

\begin{align} (1 + ax + 3x^2)\left({1 \over 2} - 2x\right)^5 & = (1 + ax + 3x^2) \left({1 \over 32} - {5 \over 8}x + 5x^2 - 20x^3 + ... \right) \\ & = ... + (1)(5x^2) + (1)(-20x^3) + ... + (ax)\left(-{5 \over 8}x\right) + (ax)(5x^2) + ... \\ & \phantom{=} + (3x^2)\left({1 \over 32}\right) + (3x^2)\left(-{5 \over 8}x\right) + ... \\ \\ & = 5x^2 - 20x^3 - {5a \over 8}x^2 + 5ax^3 + {3 \over 32}x^2 - {15 \over 8}x^3 + ... \\ & = \left(5 - {5a \over 8} + {3 \over 32} \right)x^2 + \left(-20 + 5a - {15 \over 8}\right)x^3 + ... \\ \\ \text{Since coefficient } & \text{of } x^2 \text{ is } {13 \over 2}, \\ {13 \over 2} & = 5 - {5a \over 8} + {3 \over 32} \\ {13 \over 2} - 5 - {3 \over 32} & = -{5a \over 8} \\ {45 \over 32} & = -{5a \over 8} \\ {45 \over 32} & = -{5 \over 8}a \\ \\ a & = {45 \over 32} \div -{5 \over 8} \\ & = - {9 \over 4} \\ \\ \text{Coefficient of } x^3 & = -20 + 5a - {15 \over 8} \\ & = -20 + 5\left(-{9 \over 4}\right) - {15 \over 8} \\ & = -{265 \over 8} \end{align}

(i)

\begin{align} \text{General term } & = \binom{8}{r} (x)^{8 - r} (my)^r \\ & = \binom{8}{r} x^{8 - r} (m \times y)^r \\ & = \binom{8}{r} x^{8 - r} (m)^r (y)^r \\ & = \binom{8}{r} (m)^r x^{8-r} y^r \end{align}

(ii)(a)

\begin{align} T_{5 + 1} & = \binom{8}{5} (m)^5 x^{8 - 5} y^5 \\ & = (56)(m^5) x^3 y^5 \\ & = (56)(2^5) x^3 y^5 \phantom{00000000} [m = 2] \\ & = (56)(32) x^3 y^5 \\ & = 1792x^3y^5 \\ \\ \therefore \text{Coefficient of } x^3y^5 & = 1792 \end{align}

(ii)(b)

\begin{align} \require{cancel} \\ \text{Compare } & \binom{8}{r} (m)^r x^{8-r} y^r \text{ with } -1512x^5y^3, \\ r & = 3 \\ \\ T_{3 + 1} & = \binom{8}{3}(m)^3 x^5 y^3 \\ -1512\cancel{x^5y^3} & = 56m^3 \cancel{x^5 y^3} \\ -1512 & = 56m^3 \\ \\ m^3 & = -{1512 \over 56} \\ & = -27 \\ m & = \sqrt[3]{-27} \\ & = -3 \end{align}

(i)

\begin{align} \text{General term, } \phantom{0} T_{r+1} & = \binom{n}{r}a^{n-r}b^r \\ & = \binom{10}{r} (x^3)^{10 - r} \left(-{2 \over x^2}\right)^{r} \\ & = \binom{10}{r} (x^{3(10 - r)})\left( -2 \times {1 \over x^2} \right)^r \\ & = \binom{10}{r} (x^{30 - 3r}) (-2)^r \left(1 \over x^2\right)^r \\ & = \binom{10}{r} (x^{30 - 3r}) (-2)^r (x^{-2})^r \\ & = \binom{10}{r} (x^{30 - 3r}) (-2)^r (x^{-2r}) \\ & = \binom{10}{r} (-2)^r (x^{30 - 3r})(x^{-2r}) \\ & = \binom{10}{r} (-2)^r \left[ x^{30 - 3r + (-2r)} \right] \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ & = \binom{10}{r} (-2)^r (x^{30 - 5r}) \end{align}
\begin{align} \text{Let } 10 & = 30 - 5r \\ 5r & = 30 - 10 \\ & = 20 \\ r & = {20 \over 5} \\ & = 4 \\ \\ T_{4 + 1} & = \binom{10}{4} (-2)^4 (x^{30 - 5(4)}) \\ & = (210)(16)(x^{10}) \\ & = 3 \phantom{.} 360x^{10} \end{align}

(ii)

\begin{align} \text{Let } -5 & = 30 - 5r \\ 5r & = 30 + 5 \\ 5r & = 35 \\ r & = {35 \over 5} \\ & = 7 \\ T_{7 + 1} & = \binom{10}{7} (-2)^7 (x^{30 - 5(7)}) \\ & = (120)(-128)(x^{-5}) \\ & = -15 \phantom{.} 360 \left(1 \over x^5\right) \\ \\ \therefore \text{Coefficient of } {1 \over x^5} & = -15 \phantom{.} 360 \end{align}

(iii)

\begin{align} \text{Let } 0 & = 30 - 5r \\ 5r & = 30 \\ r & = {30 \over 5} \\ & = 6 \\ \\ T_{6 + 1} & = \binom{10}{6}(-2)^6 (x^{30 - 5(6)}) \\ & = (210)(64)(x^0) \\ & = 13 \phantom{.} 440(1) \\ & = 13 \phantom{.} 440 \end{align}

(i)

\begin{align} \text{General term, } \phantom{0} T_{r+1} & = \binom{n}{r}a^{n-r}b^r \\ & = \binom{9}{r} (x)^{9 - r} \left(k \over x\right)^r \\ & = \binom{9}{r} (x^{9 - r}) \left( k \times {1 \over x} \right)^r \\ & = \binom{9}{r} (x^{9 - r}) (k)^r \left(1 \over x\right)^r \\ & = \binom{9}{r} (x^{9 - r}) (k)^r (x^{-1})^r \\ & = \binom{9}{r} (x^{9 - r}) (k)^r (x^{-r}) \\ & = \binom{9}{r} (k)^r (x^{9 - r})(x^{-r}) \\ & = \binom{9}{r} (k)^r \left[ x^{9 - r + (-r)} \right] \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ & = \binom{9}{r} (k)^r (x^{9 - 2r}) \\ \\ \text{Let } 3 & = 9 - 2r \\ 2r & = 6 \\ r & = 3 \\ \\ T_{3 + 1} & = \binom{9}{3} (k)^3 (x^{9 - 2(3)}) \\ & = (84)(k^3)(x^3) \\ & = 84k^3x^3 \\ \\ \therefore \text{Coefficient of } x^3 & = 84k^3 \\ \\ \\ \text{Let } 1 & = 9 - 2r \\ 2r & = 8 \\ r & = 4 \\ \\ T_{4 + 1} & = \binom{9}{4} (k)^4 (x^{9 - 2(4)}) \\ & = (126)(k^4)(x) \\ & = 126k^4x \\ \\ \therefore \text{Coefficient of } x & = 126k^4 \\ \\ \\ \text{Coefficient of } x^3 & = \text{Coefficient of } x \\ 84k^3 & = 126k^4 \\ 84 & = 126k \\ \\ k & = {84 \over 126} \\ & = {2 \over 3} \end{align}

(ii) To find the term in $x^5$, we need to find the terms in $x^3$ and $x^5$ from $\left(x + {k \over x}\right)^9$

\begin{align} \text{From (i), } T_{3 + 1} & = 84k^3 x^3 \\ & = 84 \left(2 \over 3\right)^3 x^3 \\ & = {224 \over 9}x^3 \\ \\ \text{Let } 5 & = 9 - 2r \\ 2r & = 4 \\ r & = 2 \\ \\ T_{2 + 1} & = \binom{9}{2} (k)^2 (x^{9 - 2(2)}) \\ & = (36)\left(2 \over 3\right)^2 (x^5) \\ & = (36)\left(4 \over 9\right)(x^5) \\ & = 16x^5 \\ \\ \\ (1 - 6x^2)\left(x + {k \over x}\right)^9 & = (1 - 6x^2)\left( ... + 16x^5 + {224 \over 9}x^3 + ...\right) \\ & = (1)(16x^5) + (-6x^2)\left({224 \over 9}x^3\right) + ... \\ & = 16x^5 - {448 \over 3}x^5 + ... \\ & = -{400 \over 3}x^5 + ... \\ \\ \therefore \text{Coefficient of } x^5 & = -{400 \over 3} \end{align}

(i)

\begin{align} \left(2 - {x \over 2}\right)^n & = (2)^n + \binom{n}{1} (2)^{n - 1} \left(-{x \over 2}\right)^1 + \binom{n}{2} (2)^{n - 2} \left(- {x \over 2}\right)^2 + ... \\ & = 2^n + \binom{n}{1} \left(2^n \over 2^1\right) \left(-{x \over 2}\right) + \binom{n}{2} \left(2^n \over 2^2\right)\left(x^2 \over 4\right) + ... \phantom{000000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ & = 2^n + (n)\left(2^n \over 2\right)\left(-{x \over 2}\right) + \left[ n(n - 1) \over 2 \right]\left(2^n \over 4\right)\left(x^2 \over 4\right) + ... \\ & = 2^n - n \left(2^n \over 4\right)x + n(n - 1) \left(2^n \over 32\right)x^2 + ... \\ & = 2^n - n \left(2^n \over 2^2\right)x + n(n - 1) \left(2^n \over 2^5\right)x^2 + ... \\ & = 2^n - n(2^{n - 2})x + n(n - 1)(2^{n - 5})x^2 + ... \end{align}

(ii)

\begin{align} (1 + 2x)\left(2 - {x \over 2}\right)^n & = (1 + 2x)\left[ 2^n - n(2^{n - 2})x + n(n - 1)(2^{n - 5})x^2 + ... \right] \\ & = (1)(2^n) + (2x)(2^n) + (1)[-n(2^{n - 2})x] + (2x)[-n(2^{n - 2})x] + (1)[n(n - 1)(2^{n - 5})x^2] + ... \\ & = 2^n + 2(2^n)x - n(2^{n - 2})x - 2n(2^{n - 2})x^2 + n(n - 1)(2^{n - 5})x^2 + ... \\ & = 2^n + [2(2^n) - n(2^{n - 2})]x + [-2n(2^{n - 2}) + n(n - 1)(2^{n - 5})]x^2 + ... \\ \\ \therefore a + bx^2 & = 2^n + [2(2^n) - n(2^{n - 2})]x + [-2n(2^{n - 2}) + n(n - 1)(2^{n - 5})]x^2 + ... \\ a + 0x + bx^2 & = 2^n + [2(2^n) - n(2^{n - 2})]x + [-2n(2^{n - 2}) + n(n - 1)(2^{n - 5})]x^2 + ... \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 0 & = 2(2^n) - n(2^{n - 2}) \\ n(2^{n - 2}) & = 2(2^n) \\ n(2^{n - 2}) & = (2^1)(2^n) \\ n(2^{n - 2}) & = 2^{1 + n} \\ n & = {2^{1 + n} \over 2^{n - 2}} \\ n & = 2^{1 + n - (n - 2)} \\ n & = 2^{1 + n - n + 2} \\ n & = 2^3 \\ n & = 8 \end{align}

(iii)

\begin{align} \text{When } & n = 8, \\ a + 0x + bx^2 & = 2^n + [2(2^n) - n(2^{n - 2})]x + [-2n(2^{n - 2}) + n(n - 1)(2^{n - 5})]x^2 + ... \\ a + 0x + bx^2 & = 2^8 + [2(2^8) - 8(2^6)]x + [-2(8)(2^6) + 8(7)(2^3)]x^2 \\ a + 0x + bx^2 & = 256 + 0x - 576x^2 \\ \\ \text{Comparing } & \text{the constant term,} \\ a & = 256 \\ \\ \text{Comparing } & \text{the coefficient of } x^2, \\ b & = -576 \\ \\ \therefore a & = 256, b = -576 \end{align}

To find the term in x⁸, we need to find the terms in x⁸ and x⁷ from (a - bx)x¹².

\begin{align} \text{General term } & = \binom{12}{r} (a)^{12 - r} (-bx)^{r} \\ & = \binom{12}{r} (a)^{12 - r} (-b \times x )^r \\ & = \binom{12}{r} (a)^{12 - r} (-b)^r (x)^r \\ & = \binom{12}{r} (a)^{12 - r} (-b)^r x^r \\ \\ T_{8 + 1} & = \binom{12}{8} (a)^{4} (-b)^8 x^8 \\ & = (495) (a^4)(b^8)x^8 \\ & = 495a^4b^8 x^8 \\ \\ T_{7 + 1} & = \binom{12}{7} (a)^5 (-b)^7 x^7 \\ & = (792)(a^5)(-b^7)x^7 \\ & = -792a^5 b^7 x^7 \\ \end{align} \begin{align} \require{cancel} (1 + x)(a - bx)^{12} & = (1 + x)(... - 792a^5b^7x^7 + 495a^4b^8 x^8 + ...) \\ & = ... + (1)(495a^4b^8x^8) + (x)(-792a^5b^7x^7) + ... \\ & = 495a^4b^8 - 792a^5b^7 x^8 + ... \\ & = (495a^4b^8 - 792a^5b^7)x^8 + ... \\ \\ \therefore \text{Coefficient of } x^8 & = 495a^4b^8 - 792a^5b^7 \\ 0 & = 495a^4b^8 - 792a^5b^7 \\ 792a^5b^7 & = 495a^4b^8 \\ a^5b^7 & = {495 \over 792}a^4b^8 \\ a^5b^7 & = {5 \over 8}a^4b^8 \\ {a^\cancel{5}\cancel{b^7} \over \cancel{a^4}b^\cancel{8}} & = {5 \over 8} \\ {a \over b} & = {5 \over 8} \end{align}

(i)

\begin{align} (1 + 3x)^n & = (1)^n + \binom{n}{1}(1)^{n - 1}(3x)^1 + \binom{n}{2}(1)^{n - 2}(3x)^2 + \binom{n}{3}(1)^{n - 3}(3x)^3 + ... \\ & = ... + \binom{n}{2}(1)^{n - 2}(3x)^2 + \binom{n}{3}(1)^{n - 3}(3x)^3 + ... \\ & = ... + \binom{n}{2}(1)(9x^2) + \binom{n}{3}(1)(27x^3) + ... \\ & = ... + 9\binom{n}{2}x^2 + 27\binom{n}{3}x^3 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = 9\binom{n}{2} \\ \text{Coefficient of } x^3 & = 27\binom{n}{3} \\ \\ 6 \times \text{Coefficient of } x^2 & = \text{Coefficient of } x^3 \\ 6 \times 9\binom{n}{2} & = 27\binom{n}{3} \\ 54\binom{n}{2} & = 27\binom{n}{3} \\ {54 \over 27} \binom{n}{2} & = \binom{n}{3} \\ \\ 2\binom{n}{2} & = \binom{n}{3} \phantom{000} \text{(Shown)} \end{align}

(ii)

\begin{align} \binom{n}{3} & = 2\binom{n}{2} \\ {n(n - 1)(n - 2) \over 6} & = 2\left[ n(n - 1) \over 2\right] \\ {n(n - 1)(n - 2) \over 6} & = n(n - 1) \\ {n - 2 \over 6} & = {\cancel{n(n - 1)} \over \cancel{n(n - 1)}} \\ {n - 2 \over 6} & = 1 \\ n - 2 & = 6 \\ n & = 8 \end{align}

(i)

\begin{align} (2 + p)^5 & = (2)^5 + {5 \choose 1} (2)^4 (p) + {5 \choose 2} (2)^3 (p)^2 + {5 \choose 3} (2)^2 (p)^3 + ... \\ & = 32 + (5)(16)(p) + (10)(8)(p^2) + (10)(4)(p^3) + ... \\ & = 32 + 80p + 80p^2 + 40p^3 + ... \end{align}

(ii)

\begin{align} (2 + p)^5 & = 32 + 80p + 80p^2 + 40p^3 + ... \\ \\ \text{Replace } & p \text{ by } x - 2x^2, \\ (2 + x - 2x^2)^5 & = 32 + 80(x - 2x^2) + 80(x - 2x^2)^2 + 40(x - 2x^2)^3 + ... \\ & = 32 + 80x - 160x^2 + 80[x(1 - 2x)]^2 + 40[x(1 - 2x)]^3 + ... \\ & = 32 + 80x - 160x^2 + 80x^2(1 - 2x)^2 + 40x^3 (1 - 2x)^3 + ... \\ & = 32 + 80x - 160x^2 + 80x^2(1 - 4x + 4x^2) + 40x^3 (1 - ...) + ... \\ & = 32 + 80x - 160x^2 + 80x^2 - 320x^3 + ... + 40x^3 + ... \\ & = 32 + 80x - 80x^2 - 280x^3 + ... \end{align}

\begin{align} (a + bx + cx^2)^4 & = [a + x(b + cx)]^4 \\ & = (a)^4 + {4 \choose 1} (a)^3 [x(b + cx)] + {4 \choose 2} (a)^2 [x(b + cx)]^2 + {4 \choose 3} (a) [x(b + cx)]^3 + [x(b + cx)]^4 \\ & = a^4 + (4)(a^3)(bx + cx^2) + (6)(a^2)(x^2)(b + cx)^2 + (4)(a)(x^3)(b + cx)^3 + ... \\ & = a^4 + 4a^3(bx + cx^2) + 6a^2 x^2 (b^2 + 2bcx + c^2 x^2) + 4ax^3(b^3 + ...) + ... \\ & = a^4 + 4a^3bx + 4a^3cx^2 + 6a^2 b^2 x^2 + 12a^2bcx^3 + ... + 4a b^3 x^3 + ... \\ & = a^4 + 4a^3bx + (4a^3c + 6a^2 b^2)x^2 + (12a^2 bc + 4a b^3)x^3 + ... \\ \\ \text{Comparing } & \text{constants,} \\ a^4 & = 81 \\ a & = \pm \sqrt[4]{81} \\ a & = 3 \text{ or } - 3 \text{ (Reject, since } a > 0) \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 4a^3b & = 216 \\ a^3b & = 54 \\ (3)^3 b & = 54 \\ 27b & = 54 \\ b & = 2 \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 4a^3 c + 6a^2 b^2 & = 108 \\ 4(3)^3 c + 6(3)^2 (2)^2 & = 108 \\ 108c + 216 & = 108 \\ 108c & = -108 \\ c & = -1 \\ \\ \text{Comparing } & \text{coefficients of } x^3, \\ 12a^2bc + 4ab^3 & = d \\ 12(3)^2(2)(-1) + 4(3)(2)^3 & = d \\ -120 & = d \\ \\ \therefore a & = 3, b = 2, c = -1, d = -120 \end{align}

(i) Term independent of x refers to the term with x to the power of 0

\begin{align} T_{r + 1} & = {n \choose r} (2x^2)^{n - r} \left(-{1 \over \sqrt{x}}\right)^r \\ & = {n \choose r}(2)^{n - r} (x^2)^{n - r} (-1)^r (x^{-{1 \over 2}})^r \\ & = {n \choose r} (2)^{n - r} (x)^{2n - 2r} (-1)^r (x)^{-{1 \over 2}r} \\ & = {n \choose r} (2)^{n - r} (-1)^r x^{2n - 2{1 \over 2}r} \\ \\ 2n - 2{1 \over 2}r & = 0 \\ 2n & = 2{1 \over 2}r \\ n & = {5 \over 4}r \\ \\ \text{Since } & r \text{ is an integer,} \\ \text{Smallest } n & = {5 \over 4}(4) \\ & = 5 \end{align}

(ii)

\begin{align} T_{r + 1} & = {n \choose r} (2)^{n - r} (-1)^r x^{2n - 2{1 \over 2}r} \\ & = {5 \choose r} (2)^{5 - r} (-1)^r x^{10 - 2{1 \over 2}r} \\ \\ x^7 \sqrt{x} & = x^7 x^{1 \over 2} \\ & = x^{7{1 \over 2}} \\ \\ 10 - 2{1 \over 2}r & = 7{1 \over 2} \\ -2{1 \over 2}r & = -2{1 \over 2} \\ r & = 1 \\ \\ T_{1 + 1} & = {5 \choose 1} (2)^{5 - 1} (-1)^1 x^{10 - 2{1 \over 2}(1)} \\ & = (5)(16)(-1) x^{7{1 \over 2}} \\ & = -80 x^{7{1 \over 2}} \\ \\ \therefore \text{Required } & \text{coefficient} = -80 \end{align}

(i)

\begin{align} (a + b)^n & = (a)^n + {n \choose 1} (a)^{n - 1} (b) + {n \choose 2} (a)^{n - 2} (b)^2 + {n \choose 3} (a)^{n - 3} (b)^3 + ... \\ & = ... + (n)(a^{n - 1})(b) + {n(n - 1) \over 2} (a^{n - 2})(b^2) + {n(n - 1)(n - 2) \over 6} (a^{n - 3})(b^3) + ... \\ \\ p & = (n)(a^{n - 1})(b) \\ \\ q & = {n(n - 1) \over 2} (a^{n - 2})(b^2) \\ \\ r & = {n(n - 1)(n - 2) \over 6} (a^{n - 3})(b^3) \\ \\ \\ pr & = (n)(a^{n - 1})(b) \left[{n(n - 1)(n - 2) \over 6}\right] (a^{n - 3})(b^3) \\ & = {n^2 (n - 1)(n - 2) \over 6} (a^{2n - 4})(b^4) \\ & = {n^2 (n - 1)(n - 2)(a^{2n - 4})(b^4) \over 6} \\ \\ q^2 & = \left[ {n(n - 1) \over 2} (a^{n - 2})(b^2) \right]^2 \\ & = {n^2 (n - 1)^2 \over 4} (a^{2n - 4})(b^4) \\ & = {n^2 (n - 1)^2 (a^{2n - 4})(b^4) \over 4} \\ \\ {pr \over q^2} & = pr \div q^2 \\ & = {n^2 (n - 1)(n - 2)(a^{2n - 4})(b^4) \over 6} \div {n^2 (n - 1)^2 (a^{2n - 4})(b^4) \over 4} \\ & = {n^2 (n - 1)(n - 2)(a^{2n - 4})(b^4) \over 6} \times {4 \over n^2(n - 1)^2 (a^{2n- 4})(b^4)} \\ & = {4n^2 (n - 1)(n - 2)(a^{2n - 4})(b^4) \over 6n^2(n - 1)^2 (a^{2n- 4})(b^4)} \\ & = {2(n - 2) \over 3(n - 1)} \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {pr \over q^2} & = {2(n - 2) \over 3(n - 1)} \\ \\ \text{When } p = 8, & \phantom{.} q = 24, r =36, \\ \\ {(8)(36) \over (24)^2} & = {2(n - 2) \over 3(n - 1)} \\ {1 \over 2} & = {2(n - 2) \over 3(n - 1)} \\ 3(n - 1) & = 2[2(n - 2)] \\ 3n - 3 & = 4(n - 2) \\ 3n - 3 & = 4n - 8 \\ -n & = -5 \\ n & = 5 \end{align}

\begin{align} a^3 & = \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3 \\ & = \left( \sqrt[3]{2 + \sqrt{5}} \right)^3 + {3 \choose 1} \left( \sqrt[3]{2 + \sqrt{5}} \right)^2 \left( \sqrt[3]{2 - \sqrt{5}} \right) \\ & \phantom{0} + {3 \choose 2} \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right)^2 + \left( \sqrt[3]{2 - \sqrt{5}} \right)^3 \\ \\ & = 2 + \sqrt{5} + (3)\left( \sqrt[3]{2 + \sqrt{5}} \right)^2 \left( \sqrt[3]{2 - \sqrt{5}} \right) \\ & \phantom{0} + (3) \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right)^2 + 2 - \sqrt{5} \\ \\ & = 4 + 3\left( \sqrt[3]{2 + \sqrt{5}} \right)^2 \left( \sqrt[3]{2 - \sqrt{5}} \right) + 3 \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right)^2 \\ & = 4 + 3 \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right) \left[ \left( \sqrt[3]{2 + \sqrt{5}} \right) + \left( \sqrt[3]{2 - \sqrt{5}}\right) \right] \\ & = 4 + 3 \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right) (a) \\ & = 4 + 3 a \left( \sqrt[3]{2 + \sqrt{5}} \right) \left( \sqrt[3]{2 - \sqrt{5}} \right) \\ & = 4 + 3 a \sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})} \\ & = 4 + 3 a [(2)^2 - (\sqrt{5})^2] \\ & = 4 + 3 a (4 - 5) \\ & = 4 + 3 a (-1) \\ & = 4 - 3a \phantom{00} \text{ (Shown)} \end{align}
\begin{align} a^3 + & \phantom{.} 3a - 4 = 0 \\ \\ \text{Let } f(a) & = a^3 + 3a - 4 \\ \\ f(1) & = (1)^3 + 3(1) - 4 \\ & = 0 \\ \\ \text{By Factor theo} & \text{rem, } a - 1 \text{ is a factor of } f(a) \\ \\ a^3 + 3a - 4 & = (a - 1)(a^2 + ba + c) \\ & = a^3 + ba^2 + ca - a^2 - ba - c \\ & = a^3 + (b - 1)a^2 + (c - b)a - c \\ \\ \text{Comparing } & \text{coefficients of } a^2, \\ 0 & = b - 1 \\ 1 & = b \\ \\ \text{Comparing } & \text{constants,} \\ -4 & = -c \\ 4 & = c \\ \\ \\ (a - 1)&(a^2 + a + 4) = 0 \end{align} \begin{align} a - 1 & = 0 && \text{ or } & a^2 + a + 4 & = 0 \\ a & = 1 &&& b^2 - 4ac & = (1)^2 - 4(1)(4) \\ & &&& & = -15 \implies \text{No real roots} \end{align}