Chapter 7 - Binomial Theorem

Index $n=8$ is even. Middle term is $T_{8/2+1} = T_5$.
$T_5 = {^8C_4} (\frac{p}{2})^{8-4} (2)^4 = 70 \cdot \frac{p^4}{16} \cdot 16 = 70p^4$.
Given $70p^4 = 1120 \implies p^4 = 16 \implies p = \pm 2$.

Index is $2n$ (even). Middle term is $T_{n+1}$.
$T_{n+1} = {^{2n}C_n} (x)^n (-\frac{1}{x})^n = {^{2n}C_n} (-1)^n$.
${^{2n}C_n} = \frac{(2n)!}{(n!)^2} = \frac{1 \cdot 2 \cdot 3 \dots (2n)}{n! n!} = \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] [2 \cdot 4 \cdot 6 \dots (2n)]}{n! n!}$
$= \frac{[1 \cdot 3 \dots (2n-1)] 2^n [1 \cdot 2 \dots n]}{n! n!} = \frac{1 \cdot 3 \dots (2n-1)}{n!} 2^n$.
Multiplying by $(-1)^n$, we get $\frac{1 \cdot 3 \dots (2n-1)}{n!} (-2)^n$.

$T_7$ (beginning) $= {^nC_6} (2^{1/3})^{n-6} (3^{-1/3})^6$.
$T_7$ (end) $= {^nC_6} (3^{-1/3})^{n-6} (2^{1/3})^6$.
Ratio $= \frac{2^{(n-6)/3} 3^{-2}}{3^{-(n-6)/3} 2^2} = 2^{\frac{n-6}{3}-2} 3^{-2+\frac{n-6}{3}} = 2^{\frac{n-12}{3}} 3^{\frac{n-12}{3}} = 6^{\frac{n-12}{3}}$.
Given ratio $= \frac{1}{6} = 6^{-1}$.
$\frac{n-12}{3} = -1 \implies n-12 = -3 \implies n=9$.

$(x+a)^n = O + E$.
$(x-a)^n = O - E$ (terms with odd powers of $a$ become negative, which are the even positioned terms $T_2, T_4...$).
(i) $O^2 - E^2 = (O+E)(O-E) = (x+a)^n (x-a)^n = (x^2 - a^2)^n$.
(ii) $4OE = (O+E)^2 - (O-E)^2 = ((x+a)^n)^2 - ((x-a)^n)^2 = (x+a)^{2n} - (x-a)^{2n}$.

General term $T_{r+1} = {^{2n}C_r} (x^2)^{2n-r} (x^{-1})^r = {^{2n}C_r} x^{4n-3r}$.
For $x^p$, $4n-3r = p \implies 3r = 4n-p \implies r = \frac{4n-p}{3}$.
Coefficient is ${^{2n}C_r} = \frac{(2n)!}{r! (2n-r)!}$.
$2n-r = 2n - \frac{4n-p}{3} = \frac{6n-4n+p}{3} = \frac{2n+p}{3}$.
Substituting $r$ and $2n-r$, we get the result.

Let the second part be expanded. General term $T_{r+1} = {^9C_r} (\frac{3}{2}x^2)^{9-r} (-\frac{1}{3x})^r = {^9C_r} (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$.
We need term independent of $x$ in $(1 + x + 2x^3) \sum T_{r+1}$.
1. $1 \times x^{18-3r} \implies 18-3r=0 \implies r=6$. Coeff $= {^9C_6} (3/2)^3 (-1/3)^6 = \frac{7}{18}$.
2. $x \times x^{18-3r} \implies 19-3r=0$ (No integer solution).
3. $2x^3 \times x^{18-3r} \implies 21-3r=0 \implies r=7$. Coeff $= 2 \times {^9C_7} (3/2)^2 (-1/3)^7 = 2 \times (-\frac{1}{54}) = -\frac{1}{27}$.
Total $= \frac{7}{18} - \frac{1}{27} = \frac{21-2}{54} = \frac{19}{54}$.
Wait, let me re-calculate $K_7$.
$K_7 = {^9C_7} (3/2)^2 (-1/3)^7 = 36 \cdot (9/4) \cdot (-1/2187) = 81 \cdot (-1/2187) = -1/27$.
Term is $2 \times K_7 = -2/27$.
Total $= 7/18 - 2/27 = (21-4)/54 = 17/54$.
Correct.

$\sum_{r=0}^{20} {^{20}C_r} = 2^{20}$.
$\sum_{r=0}^{9} {^{20}C_r} + {^{20}C_{10}} + \sum_{r=11}^{20} {^{20}C_r} = 2^{20}$.
Since ${^{20}C_r} = {^{20}C_{20-r}}$, the two sums are equal.
$2 \sum_{r=0}^{9} {^{20}C_r} + {^{20}C_{10}} = 2^{20}$.
$\sum_{r=0}^{9} {^{20}C_r} = \frac{2^{20} - {^{20}C_{10}}}{2} = 2^{19} - \frac{1}{2}{^{20}C_{10}}$.
Required sum $= \sum_{r=0}^{10} {^{20}C_r} = \sum_{r=0}^{9} {^{20}C_r} + {^{20}C_{10}} = 2^{19} - \frac{1}{2}{^{20}C_{10}} + {^{20}C_{10}} = 2^{19} + \frac{1}{2}{^{20}C_{10}}$.

$7^9 + 9^7 = (8-1)^9 + (1+8)^7 = -(1-8)^9 + (1+8)^7$.
$-(1 - 9 \cdot 8 + \binom{9}{2}8^2 - \dots) + (1 + 7 \cdot 8 + \binom{7}{2}8^2 + \dots)$.
$= -1 + 72 - 36 \cdot 64 + \dots + 1 + 56 + 21 \cdot 64 + \dots$
$= (72+56) + 64(\dots) = 128 + 64k$.
Since 128 is divisible by 64, the expression is divisible by 64.

Expression is $(2x + y^3)^{28}$. Number of terms is $28+1 = 29$.

Middle terms are $T_n$ and $T_{n+1}$. Coefficients are ${^{2n-1}C_{n-1}}$ and ${^{2n-1}C_n}$.
Sum $= {^{2n-1}C_{n-1}} + {^{2n-1}C_n} = {^{2n}C_n}$.
The statement claims it is ${^{2n-1}C_n}$, which is incorrect.

$3^{400} = (3^4)^{100} = 81^{100} = (1+80)^{100} = 1 + 100(80) + \dots = 1 + 8000 + \dots$
The last two digits are 01.

General term $x^{2n-2r} x^{-r} = x^{2n-3r}$.
For independent term, $2n=3r$. Thus $n$ must be a multiple of 3, not necessarily 2.

Number of terms is $n+1$, which is one *more* than the power $n$.