Binomial Theorem

$ \begin{aligned} &\\ &\begin{aligned} & \begin{aligned} (1 & \left.-3 x+2 x^3\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \\ T_{r+1} & ={ }^9 C_r\left(\frac{3 x^2}{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^r \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{18-3 r} \\ & 1\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9-3 x\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \\ & +2 x^3\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &r=6 \quad r=\frac{19}{3} \text { not possible } \quad r=7\\ &\Rightarrow r=6\\ &\text { We get, } 1\left(T_{6+1}\right)+2 x^3\left(T_{7+1}\right) \end{aligned} $

$ \begin{aligned} & ={ }^9 C_6\left(\frac{3 x^2}{2}\right)^3\left(\frac{-1}{3 x}\right)^6+2 x^3 \quad\left({ }^9 C_7\left(\frac{3 x^2}{2}\right)^2\left(\frac{-1}{3 x}\right)^7\right) \end{aligned} $

$ =\frac{9 \times 8 \times 7}{3!} \times \frac{(3)^3}{(2)^3}\left(\frac{1}{3}\right)^6+2 \times \frac{9 \times 8}{2}\times \frac{3^2}{2^2}\left(\frac{-1}{3}\right)^7 $

$ =\frac{7}{18}-\frac{2}{27}=\frac{17}{54} $

$ \begin{aligned} & { }^{20} C_0+{ }^{21} C_1+{ }^{22} C_2 \ldots{ }^{40} C_{20}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow{ }^{41} C_{21}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow \quad \frac{41}{21}{ }^{40} C_{20}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow \quad p^2-q^2=(41)^2-(21)^2 \\ & \quad=(41+21)(41-21)=1240 \end{aligned} $

$ \begin{aligned} & \text { } x=\frac{2 \cdot 5}{2!3}+\frac{2 \cdot 5 \cdot 7}{3!3^2}+\frac{2 \cdot 5 \cdot 7 \cdot 9}{4!3^3}+\ldots \\ & \begin{aligned} \Rightarrow \frac{x}{2}+1+\frac{1}{1!}= & 1+\frac{3}{1!}\left(\frac{1}{3}\right) \\ & +\frac{3 \cdot 5}{2!}\left(\frac{1}{3}\right)^2+\frac{3 \cdot 5 \cdot 7}{3!}\left(\frac{1}{3}\right)^3 \ldots \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{x}{2}+2=\left(1-\frac{2}{3}\right)^{-\frac{3}{2}} \\ & \Rightarrow \frac{x}{2}+2=3 \sqrt{3} \Rightarrow \frac{x^2}{4}+4+2 x=27 \\ & \Rightarrow x^2+8 x+8=27 \times 4-8=100 \end{aligned} $

$ \text { } \begin{aligned} & \frac{x}{(x-1)^2(x-2)} \\ & =x(x-1)^{-2}(x-2)^{-1} \\ & =-\frac{x}{2}(1-x)^2\left(1-\frac{x}{2}\right)^{-1} \end{aligned} $

Expansion is possible only, when $|x|<1$ which is not mentions in question.

The given expression $(7-5 x)^{-2 / 3}$ can be written as, $7^{-2 / 3}\left(1-\frac{5}{7} x\right)^{-2 / 3}$ Now, the expansion of $\left(1-\frac{5}{7} x\right)^{-2 / 3}$ is valid, if

$ \begin{aligned} & \left|\frac{5}{7} x\right| < 1 \Rightarrow \frac{5}{7}|x| < 1 \\ & \Rightarrow|x| < \frac{7}{5} \Rightarrow-\frac{7}{5} < x < \frac{7}{5} \end{aligned} $

Now, it can be concluded that $c=7 / 5$

$ \therefore \quad 5 c+7=5\left(\frac{7}{5}\right)+7=7+7=14 $

Given, $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n \therefore f(2023)$ be the coefficient of $x^{2023}$ in the expansion of $(1+x)(1-x)^{2023}$

$ \begin{aligned} & \text { Now, }(1+x)(1-x)^{2023} \\ & \begin{array}{l} =(1+x)(1-2023 x+\ldots+2023 x^{2022}-x^{2023} \\ \begin{aligned} f(2023) & \left.=1 \times(-1)+1 \times 2023 x^{2022}-x^{2023}\right) \\ & =-1+2023=2022 \end{aligned} \end{array} \end{aligned} $

$ \begin{aligned} & \text { Given, } y=\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \\ & \begin{array}{r} \Rightarrow 1+y=1+\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \\ \Rightarrow 1+y=1+\frac{3}{1}\left(\frac{1}{4}\right)^1 \\ +\frac{3 \cdot 5}{2!}\left(\frac{1}{4}\right)^2+\frac{3 \cdot 5 \cdot 7}{3!}\left(\frac{1}{4}\right)^3+\ldots \end{array} \\ & \Rightarrow 1+y=\left(1-\frac{1}{2}\right)^{-3 / 2} \\ & {\left[\begin{array}{r} \because(1-x)^{-p / q}=1+\frac{p}{1!}\left(\frac{x}{q}\right)^1+\frac{p(p+q)}{2!}\left(\frac{x}{q}\right)^2 \\ +\frac{p(p+q)(p+2 q)}{3!}+\ldots \end{array}\right]} \end{aligned} $

$ \begin{aligned} & \Rightarrow 1+y=\left(\frac{1}{2}\right)^{-3 / 2} \\ & \Rightarrow 1+y=2^{3 / 2} \Rightarrow(1+y)^2=\left(2^{3 / 2}\right)^2 \\ & \Rightarrow y^2+2 y+1=8 \Rightarrow y^2+2 y-7=0 \end{aligned} $

Given, binomial expansion

$ \begin{aligned} (2 x-3 y)^5 & =(2 x)^5\left(1-\frac{3 y}{2 x}\right)^5 \\ & =\left(2 \times \frac{3}{2}\right)^5\left(1-\frac{3 y}{2 \times 3 / 2}\right)^5 \\ & =3^5(1-y)^5 \end{aligned} $

Now, $\frac{T r+1}{T r}=\frac{n-r+1}{r}|-y|$

$ \begin{aligned} & \Rightarrow \frac{5-r+1}{r}\left|-\frac{2}{3}\right| \Rightarrow \frac{6-r}{r}\left(\frac{2}{3}\right) \\ & \Rightarrow \frac{12-2 r}{3 r} \\ & \therefore \frac{T r+1}{T r} \geq 1 \Rightarrow \frac{12-2 r}{3 r} \geq 1 \\ & \Rightarrow 12-2 r \geq 3 r \Rightarrow 12 \geq 5 r \\ & \Rightarrow r \leq \frac{12}{5} \Rightarrow r \leq 2 \frac{2}{5} \end{aligned} $

Greatest term $=T_3$

$ \begin{aligned} & =3^5 \times{ }^5 C_2(-1)^2(y)^2 \\ & =3^5 \times \frac{5 \times 4}{2 \times 1} \times\left(\frac{2}{3}\right)^2=3^3 \times 10 \times 4 \\ & =27 \times 40=1080 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \frac{2 x^3+3 x^2+3 x+5}{\left(x^2+1\right)\left(x^2+2\right)} \\ & =\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(2+x^2\right)^{-1} \\ & =\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1} \\ & \quad\left[2\left(1+\frac{x^2}{2}\right)\right]^{-1} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{2}\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1} \\ & =\frac{1}{2}\left(1+\frac{x^2}{2}\right)^{-1} \\ & \left(1-x^2+x^4-x^6+\ldots\right) \\ & \left(1-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{6}+\ldots\right) \end{aligned}\\ &\text { Now, coefficient of } x^5\\ &\begin{aligned} & =\frac{1}{2}\left[\begin{array}{l} 2 \times 1 \times\left(-\frac{1}{2}\right)+2 \times(-1) \times 1+3 \\ \times 1 \times \frac{1}{4}+3 \times 1 \times 1+3 \times(-1) \times\left(-\frac{1}{2}\right) \end{array}\right] \\ & =\frac{1}{2}\left[-1-2+\frac{3}{4}+3+\frac{3}{2}\right]=\frac{1}{2}\left(\frac{9}{4}\right)=\frac{9}{8} \end{aligned} \end{aligned} $

Number of terms in expansion of

$ \begin{aligned} (x-2 y+3 z)^5 & ={ }^{5+3-1} C_{3-1} \\ P & ={ }^7 C_2 \Rightarrow P=21 \end{aligned} $

Coefficient of $x^p y^q z^r$ in $(a x+b y+c z)^n$ is

$ \frac{n!}{p!q!r!}\left(a^p \cdot b^q \cdot c^r\right) $

$q=$ Coefficient of $x^2 y z^2$ in $(x-2 y+3 z)^5$

$ =\frac{5!}{2!\times 1!\times 2!}\left\{1^2 \cdot(-2)^1(3)^2\right\}=-540 $

Now, $\frac{q}{p}=-\frac{540}{21}=-\frac{180}{7}$

$ \begin{aligned} & \text { } \begin{aligned} S_{n+1}=5 \cdot C_0+8 \cdot & C_1+11 \cdot C_2 \\ & +\ldots+(n+1) \text { terms } \\ = & \sum_{r=0}^n(3 r+5) \cdot C_r \\ = & \sum_{r=0}^n\left(3 \cdot r \cdot C_r\right)+\sum_{r=0}^n 5 \cdot C_r \\ = & 3 \sum_{r=0}^n r C_r+5 \sum_{r=0}^n C_r \\ S_{n+1}= & 3\left(n 2^{n-1}\right)+5 \cdot 2^n \end{aligned} \end{aligned} $

On putting $n=10$

$ S_{11}=3\left(10 \cdot 2^9\right)+5 \cdot 2^{10}=20480 $

$\frac{1}{\sqrt{4-x}(2+x)^3}$

Using binomial expansion for negative index

$ \begin{aligned} & \frac{1}{16}\left(1-\frac{x}{4}\right)^{-1 / 2}\left(1+\frac{x}{2}\right)^{-3} \\ = & \frac{1}{16}\left(1+\frac{x}{8}+\frac{3}{128} x^2\right)\left(1-\frac{3 x}{2}+\frac{3}{2} x^2\right) \\ = & \frac{1}{16}\left(1-\frac{3 x}{2}+\frac{3}{2} x^2+\frac{x}{8}-\frac{3 x^2}{16}+\frac{3 x^2}{128}\right) \\ = & \frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right) \end{aligned} $

To determine the number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$, consider the general term in the expansion:

$ T_{r+1} = \binom{256}{r} (\sqrt{3})^{256-r} (\sqrt[8]{5})^r $

This expression simplifies to:

$ T_{r+1} = \binom{256}{r} (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}} $

For the term to be an integer, both $\frac{256-r}{2}$ and $\frac{r}{8}$ must be integers. Therefore:

$256 - r$ must be even, implying $r$ must be even.

$\frac{r}{8}$ also needs to be an integer, meaning $r$ must be a multiple of 8.

Thus, $r$ can be 0, 8, 16, 24, …, up to 256. This sequence forms an arithmetic progression (AP) where $r = 0, 8, 16, \ldots, 256$.

To find the total number of terms in this sequence:

The first term ($a$) is 0.

The common difference ($d$) is 8.

The last term ($l$) is 256.

Using the formula for the nth term of an AP, $l = a + (n-1) \cdot d$:

$ 256 = 0 + (n-1) \cdot 8 $

Solving for $n$:

$ 256 = 8(n-1) $

$ n-1 = \frac{256}{8} = 32 $

$ n = 32 + 1 = 33 $

Therefore, the number of integral terms in the expansion is 33.

We know that the expansion $(1+x)^n$ in power of $x$ is valid if $|x|<1$, where $n$ is a rational number.

Now, $1+x+x^2=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$

$ \begin{aligned} & =\left(x+\frac{1}{2}\right)^2+\frac{3}{4}-1+1 \\ & =\left[\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right]+1 \end{aligned} $

Now, the expansion of $\left(1+x+x^2\right)^{-\frac{3}{2}}$ in power of $x$ is valid, if

$ \begin{aligned} & \left|\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right|<1 \\ \Rightarrow & -1<\left(x+\frac{1}{2}\right)^2-\frac{1}{4}<1 \\ \Rightarrow & -\frac{3}{4}<\left(x+\frac{1}{2}\right)^2<\frac{5}{4} \\ \Rightarrow & 0 \leq\left(x+\frac{1}{2}\right)^2<\frac{5}{4} \\ & \quad\left[\because\left(x+\frac{1}{2}\right)^2 \geq 0 \forall x \in R\right] \\ \Rightarrow & \quad\left|x+\frac{1}{2}\right|<\frac{\sqrt{5}}{2} \end{aligned} $

$(1+x)^n=c_0+c_1 x+c_2 x^2+\ldots c_n x^n$.

By integrating on both sides, we get

$ \begin{aligned} &\begin{aligned} \frac{(1+x)^{n+1}}{n+1}=c_0 x+\frac{c_1 x^2}{2}+\frac{c_2 x^3}{3} & +\frac{c_n x^{n+1}}{n+1} \end{aligned}\\ &\begin{aligned} & \frac{2^{n+1}}{n+1}=c_0+\frac{c_1}{2}+\frac{c_2}{3}+\ldots \frac{c_n}{n+1} \\ & \therefore \quad c_0+\frac{c_1}{2}+\frac{c_2}{3}+\ldots \frac{c_n}{n+1}=\frac{2^{n+1}}{n+1} \end{aligned} \end{aligned} $

$ \begin{aligned} &\left(\sqrt{x}-\frac{k}{x^2}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(x^{\frac{1}{2}}\right)^{10-r}\left(\frac{-k}{x^2}\right)^r \\ &={ }^{10} C_r \frac{x^{\frac{10-r}{2}}}{x^{2 r}}(-k)^r \\ &={ }^{10} C_r x^{\frac{10-r}{2}-2 r}(-k)^r \\ &={ }^{10} C_r x^{\frac{10-5 r}{2}}(-k)^r \end{aligned} $

To be independent of $x, r=2$

$ \begin{array}{ll} \therefore & 405={ }^{10} C_2(-k)^2 \\ \Rightarrow & 405=\frac{10 \times 9}{2}(-k)^2 \\ \Rightarrow & 405=45 k^2 \\ \Rightarrow & 9=k^2 \\ \therefore & k= \pm 3 \end{array} $

To determine the number of rational terms in the binomial expansion $(\sqrt[4]{5} + \sqrt[5]{4})^{100}$, consider the term:

$ \sum\limits_{r=0}^{100} \binom{100}{r} \cdot 5^{\frac{100-r}{4}} \cdot 4^{\frac{r}{5}} $

For a term to be rational, both $\frac{r}{5}$ and $\frac{100-r}{4}$ must be integers.

Condition 1: $\frac{r}{5}$ must be an integer. This implies that $r$ must be a multiple of 5. The possible values of $r$ are:

$0, 5, 10, 15, 20, \ldots, 100$.

Condition 2: $\frac{100-r}{4}$ must also be an integer. For this condition, check which values from the list above also satisfy this.

For $r = 0$, $\frac{100-0}{4} = 25$ (integer)

For $r = 20$, $\frac{100-20}{4} = 20$ (integer)

For $r = 40$, $\frac{100-40}{4} = 15$ (integer)

For $r = 60$, $\frac{100-60}{4} = 10$ (integer)

For $r = 80$, $\frac{100-80}{4} = 5$ (integer)

For $r = 100$, $\frac{100-100}{4} = 0$ (integer)

The values of $r$ that make both $\frac{r}{5}$ and $\frac{100-r}{4}$ integers are $0, 20, 40, 60, 80, 100$.

Therefore, there are 6 rational terms in the expansion.

Here, $(1+x)^{101}\left(1-x+x^2\right)^{100}$

$ \begin{aligned} &=(1+x)(1+x)^{100}\left(1-x+x^2\right)^{100} \\ &=(1+x)\left(1+x^3\right)^{100} \\ &=\left(1+x^3\right)^{100}+x\left(1+x^3\right)^{100} \\ &=\Sigma^{100} C_r x^{3 r}+\Sigma^{100} C_r x^{3 r+1} \\ & 3 r=50 ; 3 r+1=50 \\ & r=\frac{50}{3} ; \quad r=\frac{49}{3} \\ & \Rightarrow \text { Coefficient of } x^{50}=0 \end{aligned} $

Given,

$\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)} $

Let, $f(r) = ({r^2} + 1)(r!)$

$ = ({r^2})(r!) + r!$

$ = r(r\,r!) + r!$

$ = r[(r + 1 - 1)r!] + r!$

$ = r[(r + 1)r! - r!] + r!$

$ = r[(r + 1)! - (r!)] + r!$

$ = r(r + 1)! - r(r!) + r!$

$ = (r + 2 - 2)(r + 1)! - r(r!) + r!$

$ = (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!$

$ = (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!$

$ = (r + 2)! - 3(r + 1)! + 2r!$

$ = [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]$

$\therefore$ $\sum\limits_{r = 1}^{20} {f(r)} $

$ = \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} } $

$ = [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]$

$ = [(22!) - (2!)] - 2[(21)! - (1!)]$

$ = 22! - 2! - 2\,.\,(21)! + 2\,.\,1!$

$ = 22! - 2\,.\,(21)!$

${(2021)^{2022}} + {(2022)^{2021}}$

$ = {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$

$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$

$ = {(7{k_2} - 1)^{674}} + (7m - 1)$

$ = (7n + 1) + (7m - 1) = 7(m + n)$ (multiple of 7)

$\therefore$ Remainder = 0

$\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } } $

$ = \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } } $

$ = {2^n}\,.\,{2^n} - {}^{2n}{C_n}$

$ = {2^{2n}} - {}^{2n}{C_n}$

${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$

For ${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$

${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}$

So, remainder is 8

and ${\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0$

So, remainder is 8

Given, Binomial expansion,

${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$

General term,

${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}$

For constant term,

${{30 - 5r} \over {24}} = 0$

$ \Rightarrow r = 6$

$\therefore$ Constant term,

${T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}$

$ = {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}$

$ = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}$

$ = 210{a^4}{b^6}$

We know, $GM \ge HM$

For terms a² and b³,

$\sqrt {{a^2}{b^3}} \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}$

$ \Rightarrow \sqrt {{a^2}{b^3}} \ge {2 \over 4}$

$ \Rightarrow {a^2}{b^3} \ge {1 \over 4}$

$ \Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}$

$\therefore$ ${a^4}{b^6} \ge {1 \over {16}}$

$\therefore$ Minimum value of ${a^4}{b^6} = {1 \over {16}}$

$\therefore$ Minimum value of constant term

${T_7} = 210 \times {a^4}{b^6}$

$ = 210 \times {1 \over {16}}$

$ = {{105} \over 8}$

Given,

${9^n} - 8n - 1 = 64\alpha $

$ \Rightarrow \alpha = {{{{(1 + 8)}^n} - 8n - 1} \over {64}}$

$ = {{\left( {{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,{8^1} + {}^n{C_2}\,.\,{8^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \right) - 8n - 1} \over {{8^2}}}$

$ = {{1 + 8n + {}^n{C_2}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^n} - 8n - 1} \over {{8^2}}}$

$ = {{{}^n{C_2}\,.\,{8^2} + {}^n{C_3}\,.\,{8^3}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \over {{8^2}}}$

$ = {}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,.....\,\,{}^n{C_n}\,.\,{8^{n - 2}}$

Also given,

${6^n} - 5n - 1 = 25\beta $

$ \Rightarrow \beta = {{{{(1 + 5)}^n} - 5n - 1} \over {25}}$

$ = {{{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,5 + {}^n{C_2}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^n} - 5n - 1} \over {{5^2}}}$

$ = {{1 + 5n + {}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^2} - 5n - 1} \over {{5^2}}}$

$ = {{{}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3} + {}^n{C_4}\,.\,{5^4}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^n}} \over {{5^2}}}$

$ = {}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}$

$\therefore$ $\alpha - \beta $

$ = \left( {{}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^{n - 2}}} \right) - \left( {{}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}} \right)$

$ = {}^n{C_3}\,.\,(8 - 5) + {}^n{C_4}\,.\,({8^2} - {5^2})\,\, + \,\,....\,\, + \,\,{}^n{C_n}({8^{n - 2}} - {5^{n - 2}})$

Given,

${\left( {3{x^2} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$

$ = {{{{(3{x^8} - 2{x^7} + 5)}^{10}}} \over {{x^{50}}}}$

Now constant term in ${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} = {x^{50}}$ term in ${(3{x^8} - 2{x^7} + 5)^{10}}$

General term in ${(3{x^8} - 2{x^7} + 5)^{10}}$ is

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3{x^8})^{{n_1}}}{( - 2{x^7})^{{n_2}}}{(5)^{{n_3}}}$

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n^1}}}{( - 2)^{{n_2}}}{(5)^{{n^3}}}\,.\,{x^{8{n_1} + 7{n_2}}}$

$\therefore$ Coefficient of ${x^{8{n_1} + 7{n_2}}}$ is

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n_1}}}{( - 2)^{{n_2}}}{(5)^{{n_3}}}$

where ${n_1} + {n_2} + {n_3} = 0$

For coefficient of x⁵⁰ :

$8{n_1} + 7{n_2} = 50$

$\therefore$ Possible values of n₁, n₂ and n₃ are

$\therefore$ Coefficient of x⁵⁰

$ = {{10!} \over {1!\,6!\,3!}}{(3)^1}{( - 2)^6}{(5)^3}$

$ = {{10 \times 9 \times 8 \times 7} \over 6} \times 3 \times {5^3} \times {2^6}$

$ = 5 \times 3 \times 8 \times 7 \times 3 \times {5^3} \times {2^6}$

$ = 7 \times {5^4} \times {3^2} \times {2^9}$

$ = {2^k}\,.\,l$

$\therefore$ $l = 7 \times {5^4} \times {3^2}$ = An odd integer

and ${2^k} = {2^9}$

$ \Rightarrow k = 9$

General term of Binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$ is

${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$

$ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}$

In the term,

$\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$

Term independent of x is when

(1) $33 - 5r = 0$

$ \Rightarrow r = {{33} \over 5}\,\, \notin $ integer

(2) $33 - 5r = -2$

$ \Rightarrow 5r = 35$

$ \Rightarrow r = 7\,\, \in $ integer

(3) $33 - 5r = - 3$

$ \Rightarrow 5r = 36$

$ \Rightarrow r = {{36} \over 5}\,\, \notin $ integer

$\therefore$ Only for r = 7 independent of x term possible.

$\therefore$ Independent of x term

$ = - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)$

$ = - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)$

$ = {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}$

$ = {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}$

$ = {{33} \over {200}}$

Given,

$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $

Now,

$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)} $

$ = \left( {{}^{31}{C_1}\,.\,{}^{31}{C_0} + {}^{31}{C_2}\,.\,{}^{31}{C_1} + {}^{31}{C_3}\,.\,{}^{31}{C_2} + \,\,......\,\, + \,\,{}^{31}{C_{31}}\,.\,{}^{31}{C_{30}}} \right)$

$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{31 - 1}} + {}^{31}{C_1}\,.\,{}^{31}{C_{31 - 2}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_{31 - 31}}} \right)$

[using ${}^n{C_r} = {}^n{C_{n - r}}$]

$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{30}} + {}^{31}{C_1}\,.\,{}^{31}{C_{29}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_0}} \right)$

$ = {}^{62}{C_{30}}$

Now, $\sum\limits_{k = 1}^{30} {{}^{30}{C_k}\,.\,{}^{30}{C_{k - 1}}} $

$ = \left( {{}^{30}{C_1}\,.\,{}^{30}{C_0} + {}^{30}{C_2}\,.\,{}^{30}{C_1} + \,\,.....\,\, + \,\,{}^{30}{C_{30}}\,.\,{}^{30}{C_{29}}} \right)$

$ = \left( {{}^{30}{C_0}\,.\,{}^{30}{C_{29}} + {}^{30}{C_1}\,.\,{}^{30}{C_{28}} + \,\,....\,\, + \,\,{}^{30}{C_{29}}\,.\,{}^{30}{C_0}} \right)$

$ = {}^{60}{C_{29}}$

$\therefore$ ${}^{60}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {29!\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {{{30!} \over {30}}\,.\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{60!} \over {30!\,31!}}\left( {{{62\,.\,61} \over {32}} - 30} \right) = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow \alpha = {{62\,.\,61} \over {32}} - 30$

$ \Rightarrow 16\alpha = {{62\,.\,61 - 30 \times 32} \over 2}$

$ \Rightarrow 16\alpha = {{2822} \over 2} = 1411$

= (2016 + 5)²⁰²³ [here 2016 is divisible by 7]

= ²⁰²³C₀ (2016)²⁰²³ + .......... + ²⁰²³C₂₀₂₂ (2016) (5)²⁰²² + ²⁰²³C₂₀₂₃ (5)²⁰²³

= 2016 [²⁰²³C₀ . (2016)²⁰²² + ....... + ²⁰²³C₂₀₂₂ . (5)²⁰²²] + (5)²⁰²³

= 2016$\lambda$ + (5)²⁰²³

= 7 $\times$ 288$\lambda$ + (5)²⁰²³

= 7K + (5)²⁰²³ ...... (1)

Now, (5)²⁰²³

= (5)²⁰²² . 5

= (5³)⁶⁷⁴ . 5

= (125)⁶⁷⁴ . 5

= (126 $-$ 1)⁶⁷⁴ . 5

= 5[⁶⁷⁴C₀ (126)⁶⁷⁴ + ......... $-$ ⁶⁷⁴C₆₇₃ (126) + ⁶⁷⁴C₆₇₄]

= 5 $\times$ 126 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 5 . 7 . 18 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 7$\lambda$ + 5

Replacing (5)²⁰²³ in equation (1) with 7$\lambda$ + 5, we get,

(2021)²⁰²³ = 7K + 7$\lambda$ + 5

= 7(K + $\lambda$) + 5

$\therefore$ Remainer = 5

Given,

${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, + $ ...... ${x^{500}}$

This is a G.P. with first term ${(5 + x)^{500}}$

Common ratio $ = {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}$ and 501 terms present.

$\therefore$ Sum $ = {{{{(5 + x)}^{500}}\left( {{{\left( {{x \over {5 + x}}} \right)}^{501}} - 1} \right)} \over {{x \over {5 + x}} - 1}}$

$ = {{{{(5 + x)}^{500}}\left( {{{{x^{501}} - {{(5 + x)}^{501}}} \over {{{(5 + x)}^{501}}}}} \right)} \over {{{x - 5 - x} \over {5 + x}}}}$

$ = {{{{{x^{501}} - {{(5 + x)}^{501}}} \over {5 + x}}} \over {{{ - 5} \over {5 + x}}}}$

$ = {1 \over 5}\left( {{{\left( {5 + x} \right)}^{501}} - {x^{501}}} \right)$

Coefficient of x¹⁰¹ in ${(5 + x)^{501}}$ is $ = {}^{501}{C_{101}}\,.\,{5^{400}}$

$\therefore$ In ${1 \over 5}\left( {{{(5 + x)}^{500}} - {x^{501}}} \right)$ coefficient of x¹⁰¹ is $ = {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}$

$ = {}^{501}{C_{101}}\,.\,{5^{399}}$

${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$

Now $K = {(1 + 2)^{10}} - {2^{10}}$

$ = {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$

$ = {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$

$ = 1 + 20 + 5120 + 6\lambda $

$ = 5136 + 6\lambda + 5$

$ = 6\mu + 5$

$\lambda ,\,\mu \in N$

$\therefore$ remainder = 5

${3^{2022}}$

$ = {({3^2})^{1011}}$

$ = {(9)^{1011}}$

$ = {(10 - 1)^{1011}}$

$ = {}^{1011}{C_0}{(10)^{1011}} + \,\,.....\,\, + \,\,{}^{1011}{C_{1010}}\,.\,{(10)^1} - {}^{1011}{C_{1011}}$

$ = 10\left[ {{}^{1011}{C_0}{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right] - 1$

$ = 10\,K - 1$

[As $10\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right]$ is multiple of 10]

$ = 10K + 5 - 5 - 1$

$ = 10K - 5 + 5 - 1$

$ = 5(2K - 1) + 4$

$\therefore$ Unit digit = 4 when divided by 5.