Chapter 7 - Binomial Theorem

The expansion of $(x+a)^n + (x-a)^n$ contains $\frac{n}{2} + 1$ terms if $n$ is even.
Here $n=100$ (even).
Number of terms = $\frac{100}{2} + 1 = 50 + 1 = 51$.

Coefficient of $(3r)^{th}$ term is ${}^{2n}C_{3r-1}$.
Coefficient of $(r+2)^{nd}$ term is ${}^{2n}C_{r+1}$.
Given ${}^{2n}C_{3r-1} = {}^{2n}C_{r+1}$.
$\implies 3r-1 = r+1$ or $(3r-1) + (r+1) = 2n$.
Case 1: $2r = 2 \implies r=1$ (Rejected as $r>1$).
Case 2: $4r = 2n \implies n=2r$.

Let the terms be $T_r$ and $T_{r+1}$. Coefficients are ${}^{24}C_{r-1}$ and ${}^{24}C_r$.
Given $\frac{{}^{24}C_{r-1}}{{}^{24}C_r} = \frac{1}{4}$.
$\frac{r}{24-r+1} = \frac{1}{4} \implies 4r = 25-r \implies 5r=25 \implies r=5$.
The terms are 5th and 6th.

Coefficient of $x^n$ in $(1+x)^{2n}$ is ${}^{2n}C_n$.
Coefficient of $x^n$ in $(1+x)^{2n-1}$ is ${}^{2n-1}C_n$.
Ratio = $\frac{{}^{2n}C_n}{{}^{2n-1}C_n} = \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!} = \frac{2n}{n} = 2$.
Ratio is 2:1.

Coefficients are ${}^nC_1, {}^nC_2, {}^nC_3$.
Since they are in A.P., $2({}^nC_2) = {}^nC_1 + {}^nC_3$.
$2 \frac{n(n-1)}{2} = n + rac{n(n-1)(n-2)}{6}$.
$n(n-1) = n(1 + \frac{(n-1)(n-2)}{6})$.
$n-1 = 1 + \frac{n^2-3n+2}{6}$.
$6(n-2) = n^2-3n+2$.
$6n-12 = n^2-3n+2 \implies n^2-9n+14=0$.
$(n-2)(n-7)=0$. Since $n>3$ for 4th term, $n=7$.

$A = {}^{2n}C_n$, $B = {}^{2n-1}C_n$.
$\frac{A}{B} = \frac{{}^{2n}C_n}{{}^{2n-1}C_n} = 2$ (as calculated in Q21).

Middle term is $T_6 = {}^{10}C_5 (1/x)^5 (x \sin x)^5 = 252 \sin^5 x$.
Given $252 \sin^5 x = \frac{63}{8}$.
$\sin^5 x = \frac{63}{8 \times 252} = \frac{1}{32} = (1/2)^5$.
$\sin x = 1/2$.
$x = n\pi + (-1)^n \frac{\pi}{6}$.

For $(1+x)^n$ where $n$ is even, the largest coefficient is ${}^nC_{n/2}$.
Here $n=30$, so largest coefficient is ${}^{30}C_{15}$.

Number of terms = ${}^{n+3-1}C_{3-1} = {}^{n+2}C_2 = \frac{(n+2)(n+1)}{2}$.

General term $T_{r+1} = {}^{16}C_r (x^2)^{16-r} (-x^{-2})^r = {}^{16}C_r (-1)^r x^{32-4r}$.
For constant term, $32-4r=0 \implies r=8$.
Term = ${}^{16}C_8 (-1)^8 = {}^{16}C_8$.

$T_7$ (beginning) = ${}^nC_6 (2^{1/3})^{n-6} (3^{-1/3})^6$.
$T_7$ (end) = ${}^nC_6 (3^{-1/3})^{n-6} (2^{1/3})^6$.
Equating them: $2^{(n-6)/3} 3^{-2} = 3^{-(n-6)/3} 2^2$.
$2^{(n-12)/3} = 3^{-(n-12)/3}$.
This implies $\frac{n-12}{3} = 0 \implies n=12$.

$T_{r+1} = {}^{10}C_r (a^{-1})^{10-r} (-2b/3)^r$.
For $b^4$, $r=4$. Power of $a$ is $-(10-4) = -6$.
Coefficient = ${}^{10}C_4 (-2/3)^4 = 210 \times \frac{16}{81} = \frac{1120}{27}$.

$n=28$ (even). Middle term is $T_{15}$.
$T_{15} = {}^{28}C_{14} (a^3)^{14} (ba)^{14} = {}^{28}C_{14} a^{42} b^{14} a^{14} = {}^{28}C_{14} a^{56} b^{14}$.

Coefficient of $x^p$ is ${}^{p+q}C_p$.
Coefficient of $x^q$ is ${}^{p+q}C_q$.
Since ${}^nC_r = {}^nC_{n-r}$, ${}^{p+q}C_p = {}^{p+q}C_{p+q-p} = {}^{p+q}C_q$.
Ratio is 1:1.

$T_{r+1} = {}^{10}C_r (x/3)^{(10-r)/2} (3/2x^2)^r$.
Power of $x$: $\frac{10-r}{2} - 2r = 0 \implies 10-r = 4r \implies 5r=10 \implies r=2$.
Position is $r+1 = 3$.

$25 = 2 \times 13 - 1$. So $25 \equiv -1 \pmod{13}$.
$25^{15} \equiv (-1)^{15} \pmod{13} \equiv -1 \pmod{13}$.
Remainder is $13-1 = 12$.