Binomial Theorem

General term of ${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$ is T_r+1.

T_r+1 = ${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$

= ${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$

For the coefficient of x⁷,

$ \Rightarrow $ 22 - 3r = 7

$ \Rightarrow $ r = 5

So coefficient of x⁷ = ${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$ ......(1)

Now General term of ${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$ is T_r+1.

T_r+1 = ${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$

= ${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$

For the coefficient of x^-7,

11 - 3r = -7

$ \Rightarrow $ r = 6

$\therefore$ Coefficient of x^-7 = ${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$

According to question,

Coefficient of x⁷ = Coefficient of x^-7

$ \Rightarrow $ ${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$ = ${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$

$ \Rightarrow $ $ab = 1$

${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$

=${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$

${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$

= ${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}$.........(1)

We can write ${t_n}$ by rearranging like this,

${t_n}$ = ${n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}$

= ${n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}$.........(2)

[as ${{}^n{C_0}}$ = ${{}^n{C_n}}$, ${{}^n{C_1}}$ = ${{}^n{C_{n - 1}}}$......]

By adding (1) and (2) we get,

$2{t_n}$ = ${n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}$

= $n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]$

= n${S_n}$

$\therefore$ ${{{t_n}} \over {{S_n}}} = {n \over 2}$

Given $\left( {1 + x} \right){\left( {1 - x} \right)^n}$

= ${\left( {1 - x} \right)^n}$ + $x{\left( {1 - x} \right)^n}$

General term of ${\left( {1 - x} \right)^n}$ = ${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$

$\therefore$ Term containing ${x^n}$ in ${\left( {1 - x} \right)^n}$ = ${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$

So coefficient of ${x^n}$ in ${\left( {1 - x} \right)^n}$ = ${}^n{C_n}.{\left( { - 1} \right)^n}$

General term of $x{\left( {1 - x} \right)^n}$ = ${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$

$\therefore$ Term containing ${x^n}$ in $x{\left( {1 - x} \right)^n}$ = ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$

So coefficient of ${x^n}$ in $x{\left( {1 - x} \right)^n}$ = ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$

$\therefore$ coefficient of ${x^n}$ in ${\left( {1 - x} \right)^n}$ + $x{\left( {1 - x} \right)^n}$

= ${}^n{C_n}.{\left( { - 1} \right)^n}$ + ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$

= ${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$

= ${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$

= ${\left( { - 1} \right)^n}\left[ {1 - n} \right]$

For ${\left( {1 + \alpha x} \right)^4}$ the middle term ${T_{{4 \over 2} + 1}}$ = ${}^4{C_2}.{\alpha ^2}{x^2}$

$\therefore$ Coefficient of middle term = ${}^4{C_2}.{\alpha ^2}$

For ${\left( {1 - \alpha x} \right)^6}$ the middle term ${T_{{6 \over 2} + 1}}$ = ${}^6{C_3}.-{\alpha ^3}{x^3}$

$\therefore$ Coefficient of middle term = ${}^6{C_3}.{-\alpha ^3}$

$\therefore$ According to question,

${}^4{C_2}.{\alpha ^2}$ = ${}^6{C_3}.{-\alpha ^3}$

$ \Rightarrow 6 = 20 \times - \alpha $

$ \Rightarrow \alpha = - {3 \over {10}}$

General term of ${\left( {1 + x} \right)^{n}}$ is (${T_{r + 1}}$) = ${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$

$\therefore$ General term of ${\left( {1 + x} \right)^{27/5}}$ = ${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$

For first negative term, ${\left( {{{27} \over 5} - r + 1} \right)}$ < 0

$ \Rightarrow r > {{27} \over 5} + 1$

$ \Rightarrow r > {{32} \over 5}$

$ \Rightarrow r > 6.4$

$\therefore$ r = 7

${T_{7 + 1}} = {T_8}$ means 8^th term is the first negative term.

General term = ${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$
= ${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$

When ${{{256 - r} \over 2}}$ is integer then ${\left( 3 \right)^{{{256 - r} \over 2}}}$ is integer.
And when ${{r \over 8}}$ is integer then ${\left( 5 \right)^{{r \over 8}}}$ is integer.

Entire general term will be integer when ${{{256 - r} \over 2}}$ and ${{r \over 8}}$ both are integer.

${{{256 - r} \over 2}}$ is integer when r = 0, 2, 4, 6, ......, 256

${{r \over 8}}$ is integer when r = 0, 8, 16 ,......., 256

Now both ${{{256 - r} \over 2}}$ and ${{r \over 8}}$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$\therefore$ 256 = 0 + (n - 1)8 using formula of AP, t_n = a + (n - 1)d

$\therefore$ n = ${{256} \over 8} + 1$ = 32 + 1 = 33

${\left( {1 + 0.0001} \right)^{10000}}$ = ${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$

= 1 + 10000${ \times {1 \over {{{10}^4}}}}$ + ${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$+......$\infty $

< 1 + 1 + ${1 \over {2!}}$ + ${1 \over {3!}}$ + ...... $\infty $ = e = 2.71828 < 3

We know, $\,{\left( {a + b} \right)^n}$ = ${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.

So put $a$ = b = 1

$\therefore$ 2ⁿ = ${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$

According to question, 2ⁿ = 4096 = 2¹²

$ \Rightarrow n = 12$

So $\,{\left( {a + b} \right)^n}$ = $\,{\left( {a + b} \right)^{12}}$

Here n = 12 is even so formula for greatest term is
${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$

For n = 12, greatest term ${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$

$\therefore$ Coefficient of the greatest term = ${}^{12}{C_6}$ = ${{12!} \over {6!6!}}$ = 924

$\,{\left( {r + 2} \right)^{th}}$ term = ${}^{2n}{C_{r+1}}{\left( x \right)^r}$

And coefficient of $\,{\left( {r + 2} \right)^{th}}$ = ${}^{2n}{C_{r+1}}$

$3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$

And coefficient of $3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}$

According to the question,
${}^{2n}{C_{r+1}}$ = ${}^{2n}{C_{3r - 1}}$

$ \Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$

[As if ${}^n{C_p} = {}^n{C_q}$ then p + q = n]

$ \Rightarrow 4r = 2n$

$ \Rightarrow n = 2r$

Here in this expansion ${\left( {1 + x} \right)^{p + q}}$

The general term = ${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$

$\therefore$ ${x^p}$ will be present in the term = ${}^{p + q}{C_p}.{\left( x \right)^p}$

So coefficient of ${x^p}$ = ${}^{p + q}{C_p}$

And ${x^q}$ will be present in the term = ${}^{p + q}{C_q}.{\left( x \right)^q}$

$\therefore$ coefficient of ${x^q}$ = ${}^{p + q}{C_q}$

We know ${}^n{C_r}$ = ${}^n{C_{n - r}}$

$\therefore$ ${}^{p + q}{C_q}$ = ${}^{p + q}{C_{\left( {p + q} \right) - q}}$ = ${}^{p + q}{C_p}$

So coefficients of ${x^p}$ and ${x^q}$ are equal.