Chapter 7 - Binomial Theorem

General term $T_{r+1} = {}^{9}C_r (\frac{3x^2}{2})^{9-r} (-\frac{1}{3x})^r = {}^{9}C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{2(9-r)-r}$.
For term independent of $x$, the exponent of $x$ must be 0.
$18 - 2r - r = 0 \implies 18 - 3r = 0 \implies r = 6$.
Term = ${}^{9}C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \times \frac{27}{8} \times \frac{1}{729} = 84 \times \frac{1}{8 \times 27} = \frac{84}{216} = \frac{7}{18}$.

General term $T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (-kx^{-2})^r = {}^{10}C_r (-k)^r x^{\frac{10-r}{2} - 2r}$.
For term free from $x$: $\frac{10-r}{2} - 2r = 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$.
Term = ${}^{10}C_2 (-k)^2 = 45k^2$.
Given $45k^2 = 405 \implies k^2 = 9 \implies k = \pm 3$.

$(1 - x)^{16} = 1 - 16x + {}^{16}C_2 x^2 - \dots$
Product = $(1 - 3x + 7x^2)(1 - 16x + \dots)$.
Coefficient of $x$ = $1(-16) + (-3)(1) = -16 - 3 = -19$.

General term $T_{r+1} = {}^{15}C_r (\frac{3x^2}{2})^{15-r} (-\frac{1}{3x})^r$.
Exponent of $x$: $2(15-r) - r = 30 - 3r$.
For independent term: $30 - 3r = 0 \implies r = 10$.
Term = ${}^{15}C_{10} (\frac{3}{2})^5 (-\frac{1}{3})^{10} = 3003 \times \frac{243}{32} \times \frac{1}{59049} = \frac{3003}{32 \times 243} = \frac{3003}{7776} = \frac{1001}{2592}$.

(i) $n=10$ (even). Middle term is $T_{10/2 + 1} = T_6$.
$T_6 = {}^{10}C_5 (\frac{x}{a})^5 (-\frac{a}{x})^5 = -{}^{10}C_5 = -252$.
(ii) $n=9$ (odd). Middle terms are $T_{(9+1)/2} = T_5$ and $T_{(9+3)/2} = T_6$.
$T_5 = {}^{9}C_4 (3x)^5 (-\frac{x^3}{6})^4 = 126 \times 243 x^5 \times \frac{x^{12}}{1296} = \frac{189}{8} x^{17}$.
$T_6 = {}^{9}C_5 (3x)^4 (-\frac{x^3}{6})^5 = -126 \times 81 x^4 \times \frac{x^{15}}{7776} = -\frac{21}{16} x^{19}$.

$(x - x^2)^{10} = x^{10}(1 - x)^{10}$.
We need the coefficient of $x^{15}$ in $x^{10}(1 - x)^{10}$, which is the coefficient of $x^5$ in $(1 - x)^{10}$.
General term of $(1-x)^{10}$ is ${}^{10}C_r (-x)^r$.
For $x^5$, $r=5$. Coefficient = ${}^{10}C_5 (-1)^5 = -252$.

General term $T_{r+1} = {}^{15}C_r (x^4)^{15-r} (-x^{-3})^r = {}^{15}C_r (-1)^r x^{60-4r-3r} = {}^{15}C_r (-1)^r x^{60-7r}$.
We need $x^{-17}$, so $60 - 7r = -17 \implies 7r = 77 \implies r = 11$.
Coefficient = ${}^{15}C_{11} (-1)^{11} = -{}^{15}C_4 = -1365$.

Binomial coefficient of 3rd term from end = Binomial coefficient of 3rd term from beginning = ${}^nC_2$.
Given ${}^nC_2 = 45 \implies \frac{n(n-1)}{2} = 45 \implies n(n-1) = 90 \implies n = 10$.
Sixth term $T_6 = {}^{10}C_5 (y^{1/2})^5 (x^{1/3})^5 = 252 y^{5/2} x^{5/3}$.

Coefficient of $T_{2r+4}$ is ${}^{18}C_{2r+3}$.
Coefficient of $T_{r-2}$ is ${}^{18}C_{r-3}$.
Given ${}^{18}C_{2r+3} = {}^{18}C_{r-3}$.
Either $2r+3 = r-3 \implies r = -6$ (Not possible).
Or $(2r+3) + (r-3) = 18 \implies 3r = 18 \implies r = 6$.

Coefficients are ${}^{2n}C_1, {}^{2n}C_2, {}^{2n}C_3$.
Since they are in A.P., $2({}^{2n}C_2) = {}^{2n}C_1 + {}^{2n}C_3$.
$2 \frac{2n(2n-1)}{2} = 2n + \frac{2n(2n-1)(2n-2)}{6}$.
$2n(2n-1) = 2n + \frac{n(2n-1)(2n-2)}{3}$.
Divide by $n$: $2(2n-1) = 2 + \frac{(2n-1)(2n-2)}{3}$.
$6(2n-1) = 6 + (2n-1)(2n-2)$.
$12n - 6 = 6 + 4n^2 - 6n + 2$.
$12n - 6 = 4n^2 - 6n + 8$.
$4n^2 - 18n + 14 = 0 \implies 2n^2 - 9n + 7 = 0$.

$(1 + x + x^2 + x^3)^{11} = (\frac{1-x^4}{1-x})^{11} = (1-x^4)^{11} (1-x)^{-11}$.
$(1 - 11x^4 + \dots) (1 + {}^{11}C_1 x + {}^{12}C_2 x^2 + {}^{13}C_3 x^3 + {}^{14}C_4 x^4 + \dots)$.
Coefficient of $x^4$ comes from:
1. $1 \times ({}^{14}C_4 x^4) \implies {}^{14}C_4 = \frac{14 \times 13 \times 12 \times 11}{24} = 1001$.
2. $(-11x^4) \times 1 \implies -11$.
Total coefficient = $1001 - 11 = 990$.