Binomial Theorem

We are given the expression $(p-q)^{14}$. We need to find the value of $\frac{p+q}{p-q}$, given that the sum of the 7th and 8th terms in its expansion is zero.

Finding the 7th and 8th Terms

In the binomial expansion of $(p-q)^{14}$, the formula for the $(r+1)$th term is: $T_{r+1} = {^{14}C_r} p^{14-r} (-q)^r$.

The 7th term, $T_7$, is $^{14}C_6 p^{8}(-q)^{6} = {^{14}C_6} p^8 q^6$.

The 8th term, $T_8$, is $^{14}C_7 p^7(-q)^7 = -{^{14}C_7} p^7 q^7$.

Setting Up the Equation

We are told that $T_7 + T_8 = 0$.

This means: ${^{14}C_6} p^8 q^6 - {^{14}C_7} p^7 q^7 = 0$.

We can factor out $p^7 q^6$ from both terms: $p^7 q^6 \left({^{14}C_6}p - {^{14}C_7}q \right) = 0$.

Because $p$ and $q$ are not zero, we must have ${^{14}C_6}p = {^{14}C_7}q$.

Solving for $\frac{p}{q}$

Divide both sides by $q$ and ${^{14}C_6}$ to get $\frac{p}{q} = \frac{^{14}C_7}{^{14}C_6}$.

Now, $^{14}C_7 = \frac{14!}{7! \cdot 7!}$ and $^{14}C_6 = \frac{14!}{6! \cdot 8!}$.

Write $\frac{^{14}C_7}{^{14}C_6}$ as $\frac{8}{7}$ (after cancelling out common terms, which are shown in the expansion above).

Express $p$ and $q$

If $\frac{p}{q} = \frac{8}{7}$, let $p = 8k$ and $q = 7k$ for some $k$.

Find $\frac{p+q}{p-q}$

Then $p+q = 8k + 7k = 15k$ and $p-q = 8k - 7k = k$.

So, $\frac{p+q}{p-q} = \frac{15k}{k} = 15$.

General term of $(x+3 y)^{13}$ is

$ \begin{aligned} & T_{r+1}={ }^{13} C_r(x)^{13-r}(3 y)^r \\ & \text { at } x=1 / 2, y=1 / 3 \end{aligned} $

So, $3 y=1$

$ \text { then, } T_{r+1}={ }^{13} C_r\left(\frac{1}{2}\right)^{13-r}(1)^r={ }^{13} C_r\left(\frac{1}{2}\right)^{13-r} $

For maximum term

$ \begin{aligned} & \begin{aligned} \left|\frac{T_{r+1}}{T_r}\right| & =\frac{{ }^{13} C_r}{{ }^{13} C_{r-1}}\left(\frac{1}{2}\right)^{13-r-(13-(r-1))} \\ & =\frac{{ }^{13} C_r}{{ }^{13} C_{r-1}}\left(\frac{1}{2}\right)^{-1}=\frac{14-r}{r} \times 2 \end{aligned} \\ & \text { put }\left|\frac{T_{r+1}}{T_r}\right|=1 \end{aligned} $

$ \Rightarrow \quad 28-2 r=r \Rightarrow r=28 / 3 \approx 9.33 $

So, maximum occurs either $r=9$ or $r=10$

for $r=9$

$ T_{10}={ }^{13} C_9\left(\frac{1}{2}\right)^4=44.6875 $

and for $r=10$

$ T_{11}={ }^{13} C_{10}\left(\frac{1}{2}\right)^3=35.75 $

Hence, the greatest term is $T_{10}$

$ T_{10}={ }^{13} C_9\left(\frac{1}{2}\right)^4 $

We have, $(2 m+1)^{2 n}=\left[(2 m+1)^2\right]^n$

$ \begin{aligned} = & \left(4 m^2+4 m+1\right)^n \\ = & (1+4 m(m+1)]^n \\ = & { }^n C_0[(4 m)(m+1)]^n \\ & \quad+{ }^n C_1[4 m(m+1)]^{n-1}+ \\ & \quad \ldots+{ }^n C_{n-1}(4 m(m+1)]^1+{ }^n C_n \\ = & { }^n C_0\left(4^n\right)(m(m+1))^n \\ & \quad+{ }^n C_1(4 m(m+1))^{n-1} \\ & \quad+\ldots+{ }^n C_{n-1}(4)(m(m+1))+1 \end{aligned} $

$\because m(m+1)$ is always even.

$ \therefore 8(k)+1 $

∴ Remainder is 1 .

$ \begin{aligned} & \text { We have, } \sum_{r=1}^{15} r^2 \frac{{ }^{15} C_r}{{ }^{15} C_{r-1}} \\ & \begin{array}{l} \because \quad \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r} \\ \Rightarrow \quad \sum\limits_{r = 1}^{15} {{r^2}} \cdot \frac{(15-r+1)}{r} \\ \Rightarrow \quad \sum\limits_{r = 1}^{15} {{r}}(16-r) \Rightarrow \sum_{r=1}^{15}\left(16 r-r^2\right) \\ \Rightarrow \quad 16\left(\frac{15}{2} \times 16\right)-\frac{15}{6}(16)(31) \\ \quad=15(16)\left(8-\frac{31}{6}\right) \\ \quad=\frac{15 \times 16 \times 17}{6} \\ \quad=5 \times 8 \times 17=680 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{gathered} \left(\frac{1}{81}\right)^n-{ }^{2 n} C_1 \frac{10}{81^n}+{ }^{2 n} C_2 \frac{10^2}{81^n}-\ldots+\frac{10^{2 n}}{81^n} \\ \Rightarrow \quad \frac{1}{81^n}\left({ }^{2 n} C_0-{ }^{2 n} C_1 10^1+{ }^{2 n} C_2 10^2\right. \left.+\ldots+{ }^{2 n} C_{2 n} 10^{2 n}\right) \\ \Rightarrow \quad \frac{1}{81^n}(1-10)^{2 n}=\frac{9^{2 n}}{9^{2 n}}=1 \end{gathered} \end{aligned} $

We have,

$ \begin{aligned} (1+x)^{\frac{27}{5}}= & 1+\frac{27}{5} x+\frac{\frac{27}{5}\left(\frac{27}{5}-1\right)}{2!} x^2 \\ & +\frac{\frac{27}{5}\left(\frac{27}{5}-1\right)\left(\frac{27}{5}-2\right)}{3!} x^3+ \end{aligned} $

the first negative term

$ \frac{\frac{27}{5}\left(\frac{27}{5}-1\right) \ldots\left(\frac{27}{5}-6\right)}{7!} x^7 $

that is 8 term.

Coefficient of $x^{10}$ in the expansion of $\left(x+\frac{2}{x}-5\right)^{12}$

Using multinomial theorem

$ =\frac{12!}{r_{1}!r_{2}!r_{3}!} x^n\left(\frac{2}{x}\right)^{r_2}(-5)^{r_3} $

for coefficient of $x^{10}$

$ r_1+r_2+r_3=12 \Rightarrow r_1-r_2=10 $

$ \begin{aligned} &\begin{array}{|c|c|c|} \hline \boldsymbol{r}_1 & \boldsymbol{r}_2 & \boldsymbol{r}_3 \\ \hline 10 & 0 & 2 \\ \hline 11 & 1 & 0 \\ \hline \end{array}\\ &\text { Coefficient }\\ &\begin{aligned} & =\frac{12!}{10!0!2!} 2^0 \cdot(-5)^2+\frac{12!}{11!1!!} 2^{!} .(-5)^0\\ & =\frac{12 \times 11}{2} \times 25+24 \\ & =1650+24=1674 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} S_1 & =\sum_{j=1}^{10} j(j-1) \cdot{ }^{10} C_j \\ & =\sum_{j=1}^{10} j(j-1) \frac{10}{j} \cdot \frac{9}{j-1} \cdot{ }^8 C_{j-2} \\ & =\sum_{j=2}^{10} 90 \cdot{ }^8 C_{j-2}=90 \cdot 2^8 \end{aligned}\\ &\text { } \begin{aligned}and\,\, S_2 & =\sum_{j=1}^{10} j \cdot{ }^{10} C_j \\ & =\sum_{j=1}^{10} j \cdot \frac{10}{j}{ }^9 C_{j-1}=10 \sum_{j=1}^{10}{ }^9 C_{j-1} \\ & =10 \cdot 2^9 \\ S_3 & =\sum_{j=1}^{10} j^{210} C_j=10 \sum_{j=1}^{10} j \cdot{ }^9 C_{j-1} \\ & =10 \cdot 11 \cdot 2^{9-1}=110 \cdot 2^8 \\ & \quad\left[\because \sum r \cdot{ }^n C_r=n \cdot 2^{n-1}\right] \\ & =55 \cdot 2^9 \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} y & =\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots+\infty \\ & =\frac{3}{1!} \cdot \frac{1}{4}+\frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2}+\frac{3 \cdot 5 \cdot 7}{3!} \cdot \frac{1}{4^3}+\ldots+\infty \end{aligned} $

Since, using binomial expansion

$ \begin{aligned} (1-x)^{-n}=1+n x+ & \frac{n(n+1)}{2!} x^2 +\frac{n(n+1)(n+2)}{3!} x^3+\ldots \end{aligned} $

Here, $n x=\frac{3}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} &\begin{aligned} & \frac{n(n+1)}{2} x^2=\frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2} \\ \Rightarrow \quad & n(n+1) x^2=\frac{15}{16} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned}\\ &\text { From Eq. (i) and (ii), we get }\\ &\begin{aligned} \frac{n(n+1) x^2}{n^2 x^2} & =\frac{15 / 16}{(3 / 4)^2} \\ & =\frac{15 / 16}{9 / 16}=\frac{5}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{lc} \Rightarrow & \frac{n+1}{n}=\frac{5}{3} \\ \Rightarrow & 3 n+3=5 n \\ \Rightarrow & n=\frac{3}{2} \end{array}\\ &\text { So, } x=\frac{3}{4} \cdot \frac{1}{n}=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}\\ &\begin{aligned} \therefore \quad y & =(1-x)^{-n}-1=\left(1-\frac{1}{2}\right)^{-3 / 2}-1 \\ & =\left(\frac{1}{2}\right)^{-3 / 2}-1=2^{3 / 2}-1=2 \sqrt{2}-1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { So, } \quad y+1=2 \sqrt{2}-1+1=2 \sqrt{2} \\ & \quad(y+1)^2=(2 \sqrt{2})^2=8 \\ & \Rightarrow y^2+2 y+1-8=0 \\ & \Rightarrow \quad y^2+2 y-7=0 \end{aligned}\\ &\text { Thus, the quadratic equation is }\\ &y^2+2 y-7=0 \end{aligned} $

We will expand $\left(1+x-x^2\right)^6$ and find the coefficients of $x^4$ and $x^6$.

Multimonial expansion states that

$ \begin{aligned} \left(a_1+a_2+a_3\right)^n & =\sum \frac{n!}{k_{1}!k_{2}!k_{3}!} \\ a_1^k a_2^{k_2} a_3^{k_3} & \end{aligned} $

where $k_1+k_2+k_3=n$

In our case, $a_1=1, a_2=x, a_3=-x^2$ and $n=6$

$ \begin{aligned} & \text { From } k_1+k_2+k_3=6 \\ & \Rightarrow \quad k_1=6-k_2-k_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

To get $x^4$, we can have different combinations of $x$ and $x^2$

$k_1$ : number of times 1 is chosen

$k_2$ : number of times $x$ is chosen

$k_3$ : number of times $x^2$ is chosen

The equation is: $k_2+2 k_3=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, if $k_3=0$, then $k_2=4$ and $k_1=2$

If $k_3=1$, then $k_2=2$ and $k_1=3$

If $k_3=2$, then, $k_2=0, k_1=4$

Now, find the coefficients for each valid combination.

Coefficient of

$ \begin{aligned} & x^4=\frac{6!}{2!4!0!}-\frac{6!}{3!2!1!}+\frac{6!}{4!0!2!} \\ & =15-60+15=-30 \end{aligned} $

Now, for $x^6, k_2+2 k_3=6$ and

$ k_1+k_2+k_3=6 $

If $k_3=0$, then $k_2=6$, and $k_1=0$

If $k_3=1$, then $k_2=4$, and $k_1=1$

If $k_3=2$ then $k_2=2$ and $k_1=2$

If $k_3=3$, then $k_2=0$, and $k_1=3$

So, coefficient of $x^6=\frac{6!}{0!6!0!}-\frac{6!}{1!4!1!}+\frac{6!}{2!2!2!}-\frac{6!}{3!0!3!}$

$ =1-30+90-20=41 $

$\therefore$ Sum of the coefficient of $x^4$ and $x^6=41-30=11$

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} 11^{12}-11^2 & =k\left(5 \times 10^9+6 \times 10^9\right. + \left.33 \times 10^8+110 \times 10^7+\ldots+33\right) \\ 11^{12}-11^2 & =(10+1)^{12}-(10+1)^2 \\ & =10^{12}+12 \times 10^{11}+66 \times 10^{10}+220 \times 10^9 \cdots+66 \times 10^2+120+1-121 \end{aligned} \end{aligned} $

$ \begin{aligned} & =10^{12}+12 \times 10^{11}+66 \times 10^{10}+220 \times 10^9 \ldots+66 \times 10^2 \\ & =2 \times 10^2\left(5 \times 10^9+6 \times 10^9+33 \times 10^8\right. \left.+110 \times 10^7+\ldots+33\right) \\ & =200\left(5 \times 10^9+6 \times 10^9+33 \times 10^8\right. \left.+110 \times 10^7+\ldots+33\right) \\ & \therefore k=200 \end{aligned} $

$ \because(1+x)^n=\sum_{k=0}^n C_k x^k$

and

$ \left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\cdots $

can be grouped as

$ \begin{gathered} \sum_{k=0}^{n / 2}(-1)^k\left(C_{2 k}+C_{2 k+1}\right) \\ \text { and }(1+i)^n=\sum_{k=0}^n C_k i^k \\ \operatorname{Re}\left[(1+i)^n\right]=C_0-C_2+C_4-C_6+\cdots \\ \operatorname{Im}\left[(1+i)^n\right]=C_1-C_3+C_5-C_7+\cdots \end{gathered} $

Adding both

$ \begin{aligned} & \operatorname{Re}\left[(1+i)^n\right]+\operatorname{Im}\left[(1+i)^n\right] \\ & =\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\ldots \end{aligned} $

Now, $(1+i)^n$

$ =(\sqrt{2})^n\left(\cos \left(\frac{n \pi}{4}\right)+i \sin \left(\frac{n \pi}{4}\right)\right) $

Hence, $\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)$

$ \begin{aligned} -\ldots & =\operatorname{Re}\left[(1+i)^n\right]+\operatorname{Im}\left[(1+i)^n\right] \\ & =2^{n / 2}\left(\cos \left(\frac{n \pi}{4}\right)+\sin \left(\frac{n \pi}{4}\right)\right) \end{aligned} $

The mean of a binomial distribution is $x$, and the variance is $5$.

We know that the mean, $\mu = n p = x$. This means $n$ times $p$ is $x$.

The formula for variance is $\sigma^2 = n p(1-p) = 5$. This means $n$ times $p$ times $(1-p)$ is $5$.

From the mean formula, $p = \frac{x}{n}$.

Substitute $p = \frac{x}{n}$ into the variance formula:

$ x\left(1-\frac{x}{n}\right) = 5 $

If we open the bracket, we get: $ x - \frac{x^2}{n} = 5 $

Take $\frac{x^2}{n}$ to the other side: $ \frac{x^2}{n} = x - 5 $

Now, solve for $n$: $ n = \frac{x^2}{x - 5} $

To get a whole number for $n$, $x-5$ must divide $x^2$ exactly. Let's check different $x$ values:

If $x = 6$: $ n = \frac{36}{1} = 36 $ $36$ is an integer.

If $x = 10$: $ n = \frac{100}{5} = 20 $ $20$ is an integer.

If $x = 30$: $ n = \frac{900}{25} = 36 $ $36$ is an integer.

So, the possible values for $x$ are: $6, 10, 30$.

As we know that, coefficient of $x^{11}$ in the expansion of $(1+x)^{11}$ is ${ }^{11} C_{11}$.

Now, we have to find coefficient of $x^{11}$ in the expansion of $\left(1+\alpha x+\beta x^2\right)(1+x)^{11}$. When we choosing 1 from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^{11}$ from $(1+x)^{11}$.

So, coefficient $={ }^{11} C_{11}=1$

Similarly, we choosing $\alpha x$ from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^{10}$ from $(1+x)^{11}=$ Coefficient $=\alpha \times{ }^{11} C_{10}=11 \alpha$ and when we choosing $\beta x^2$ from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^9$ from $(1+x)=$ Coefficient $=\beta \times{ }^{11} C_9=55 \cdot \beta$

$\therefore$ Coefficient of $x^{11}$ in the expansion of $\left(1+\alpha x+\beta x^2\right)$

$(1+x)^{11}=1+11 \alpha+55 \beta=144$ ( given )          ...(i)

Just, in similar way, we have to find coefficient of $x^{10}$ such that

$ 11+55 \alpha+165 \beta=396 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get;

$ \begin{gathered} \alpha=-2 \text { and } \beta=3 \\ \therefore \alpha^2+\beta^2=(-2)^2+(3)^2=13 \end{gathered} $

Given, $\frac{-2}{3} < x < \frac{2}{3}$

$ \Rightarrow-2<3 x<2 \Rightarrow-1<\frac{3 x}{2}<1 $

Now, $\frac{1}{3 \sqrt{2-3 x}}=\frac{1}{(2-3 x)^{1 / 3}}$

$ =\frac{1}{2^{1 / 3}}\left(1-\frac{3 x}{2}\right)^{-1 / 3} $

$ =\frac{1}{3 \sqrt{2}}\left[1+\left(\frac{1}{3}\right)\left(\frac{3 x}{2}\right)+\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)}{2 \times 1}\left(\frac{3 x}{2}\right)^2\right. +\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)}{3 \times 2 \times 1}\left(\frac{3 x}{2}\right)^3\left.+\frac{\frac{1}{3}\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)\left(\frac{1}{3}+3\right)}{4 \times 3 \times 2 \times 1}\left(\frac{3 x}{2}\right)^4+\ldots\right]$

$ \begin{aligned} &\therefore 5 \text { th term when } x=\frac{1}{2} \text { is given by }\\ &\begin{array}{r} \frac{1}{\sqrt[3]{2}}\left[\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{10}{3} \times \frac{1}{24} \times \frac{81}{16} \times \frac{1}{16}\right] \\ =\frac{1}{3 \sqrt{2}}\left[\frac{7 \times 5}{24 \times 2 \times 6}\right]=\frac{35}{768(\sqrt[3]{2})} \end{array} \end{aligned} $

General term for $\left(x+y^2\right)^{13}$

$ ={ }^{13} C_n x^{13-n} y^{2 n} $

The general term for $\left(x^2+y\right)^{14}$

$ ={ }^{14} C_{r_2} x^{2 r_2} y^{14-r_2} $

Equate the power of $x$ and $y$

$ \begin{aligned} & 13-r_1=2 r_2 \\ & 2 r_1=14-r_2 \end{aligned} $

Solving these equation, we get

$ r_1=5, r_2=4 $

The common term is $x^{13-5} y^{2(9)}=x^8 y^{10}$ The common term is $x^8 y^{10}$

$ \Rightarrow \quad r=8, s=10 $

$\alpha=1$, since there is only one common term

$ \begin{aligned} & \Rightarrow \quad \alpha(r+s)=1(8+10)=18 \\ & \therefore \text { Sum } \alpha \sum_{r, s}(r+s)=18 \end{aligned} $

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

$ \begin{gathered} \Rightarrow(1+3 x)^{-3}=1-3(3 x)+\frac{(-3)(-4)(3 x)^2}{2!} \\ +\frac{(-3)(-4)(-5)(3 x)^3}{3!}+\ldots \\ \Rightarrow(1+3 x)^{-3}=1-9 x+54 x^2-270 x^3+\ldots \\ \begin{aligned} (1+4 x & \left.-3 x^2\right)\left(1-9 x+54 x^2-270 x^3+\ldots\right) \\ = & 1\left(-270 x^3\right)+4 x\left(5 x^2\right)-3 x^2(-9 x) \\ = & -270 x^3+216 x^3+27 x^3 \\ = & -270+216+27=-27 \end{aligned} \end{gathered} $

$\Rightarrow$ The coefficient of $x^3=-27$

$9^{11}+11^9=(10-1)^{11}+(10+1)^9$

$ \begin{aligned} ={ }^{11} C_0 10^{11} & +{ }^{11} C_1 10^9+\ldots{ }^{11} C_{11} \\ & +{ }^9 C_0 10^{10}+{ }^9 C_1 10^8+\ldots+{ }^9 C_9 \end{aligned} $

Sum of last two terms of both expansion as other term have at least $10^2$ as a factor

$ (110-1)+90+1=200 $

$\therefore \quad 9^{11}+11^9=$ Multiple of

$ 100+200=\text { Multiple of } 100 $

Since $200=2 \times 10^2$

So, $9^{11}+11^9$ is divisible by $10^2$

$ \therefore \quad k=2 $

Given expansion $\left(x^{\frac{1}{2}}+y^{\frac{1}{k}}\right)^{10}$

General term $={ }^{10} C_r x^{\frac{10-r}{2}} y^{\frac{r}{k}}$

Given expansion have only 9 irrational term

$\therefore$ Expansion has only two rational term Two rational term is possible

LCM of 2 and $k$ is 10

$ \therefore k=5 $

$\left(x^2+x-2\right)^5$

Apply multinomial theorem

$ \Sigma \frac{5!}{p!q!r!}\left(x^2\right)^p(x)^q(-2)^r $

Such that $p+q+r=5$ and $p, q, r$ are non-negative integer.

$ \begin{aligned} & 2 p+q=2 \\ & (p, q, r)=(1,0,4)(0,2,3) \\ & \frac{5!}{4!}\left(-4^4+\frac{5!}{2!3!}\left(-2^3\right.\right. \\ & 5 \times 16+10(-8)=80-80=0 \end{aligned} $

We have, $P_n={ }^n C_0 \cdot{ }^n C_1{ }^n C_2 \ldots{ }^n C_n$ and

$ \begin{aligned} & P_{n+1}={ }^{n+1} C_0 \cdot{ }^{n+1} C_1{ }^{n+1} C_2 \cdots{ }^{n+1} C_{n+1} \\ & \therefore \frac{P_{n+1}}{P_n}=\frac{{ }^{n+1} C_0 \cdot{ }^{n+1} C_1{ }^{n+1} C_2 \cdots{ }^{n+1} C_{n+1}}{{ }^n C_0 \cdot{ }^n C_1 \cdot{ }^n C_2 \cdots{ }^n C_n} \end{aligned} $

$ \begin{aligned} & =\left(\frac{{ }^{n+1} C_1}{{ }^n C_0}\right)\left(\frac{{ }^{n+1} C_2}{{ }^n C_1}\right)\left(\frac{{ }^{n+1} C_3}{{ }^n C_2}\right) \cdots\left(\frac{{ }^{n+1} C_{n+1}}{{ }^n C_n}\right) \\ & =\left(\frac{n+1}{1}\right)\left(\frac{n+1}{2}\right)\left(\frac{n+1}{3}\right) \cdots\left(\frac{n+1}{n+1}\right) \\ & =\frac{(n+1)^{n+1}}{(n+1)!} \end{aligned} $

Given, $\frac{x^4+1}{\left(x^2+1\right)(x-1)}$

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

Clearly, there is no terms containing $x^3$ coefficient of $x^3=0$

The given sum can be rewritten as

$ \sum_{k=0}^n \sum_{r=0}^k C_r $

Now, changing order of summarion

$ \sum_{r=0}^n \sum_{k=r}^n C_r \quad\left[\because \sum_{r=0}^n C_r=2^n\right] $

$ \begin{aligned} & \sum_{r=0}^n C_r \sum_{k=r}^n 1=\sum_{r=0}^n C_r(n-r+1) \\ & \Rightarrow(n+1) \sum_{r=0}^n C_r-\sum_{r=0}^n r \cdot C_r \\ & \quad=(n+1) 2^n-\sum_{r=0}^n r \cdot{ }^{n-1} C_{r-1} \times \frac{n}{r} \\ & \quad=(n+1) 2^n-n \sum_{r=0}^n n-1 C_r \\ & \quad=(n+1) 2^n-n \cdot 2^{n-1} \\ & \quad=n \cdot 2^n+2^n-n \cdot 2^{n-1} \\ & \quad=2^{n-1}(n+2) \end{aligned} $

$ \begin{aligned} &\text { We have, }\left(\frac{3 x-5}{4 x^2+3}\right)^{-\frac{4}{5}}=\left(\frac{4 x^2+3}{3 x-5}\right)^{\frac{4}{5}}\\ &\begin{aligned} & =\left\{\frac{4 x^2\left(1+\frac{3}{4 x^2}\right)}{3 x\left(1-\frac{5}{3 x}\right)}\right\}^{\frac{4}{5}}=\left(\frac{4 x}{3}\right)^{\frac{4}{5}} \frac{\left(1+\frac{3}{4 x^2}\right)^{4 / 5}}{\left(1-\frac{5}{3 x}\right)^{4 / 5}} \\ & \Rightarrow\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{3}{4 x^2}\right)^{\frac{4}{5}} \cdot\left(1-\frac{5}{3 x}\right)^{-\frac{4}{5}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{5} \cdot \frac{3}{4 x^2}+\ldots\right) \\ & \left(1+\frac{5}{3 x} \cdot \frac{4}{5}+\frac{4}{5}\left(+\frac{9}{5}\right)\left(\frac{5}{3 x}\right)\right)^2 \frac{1}{2!}+\ldots \ldots \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{3}{5 x^2}+\ldots\right)\left(1+\frac{4}{3 x}+\frac{2}{x^2}\right) \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{3 x}+\left(\frac{3}{5}+2\right)^{\frac{1}{x^2}}\right) \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, }\left(1+x+2 x^2\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \\ & =\left(1+x+2 x^2\right) \\ & \left\{\sum_{r=0}^9{ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r\right\} \\ & =\left(1+x+2 x^2\right) \\ & \quad\left\{\sum_{r=0}^9{ }^9 C_r\left(\frac{3}{2}\right)^{9-r} x^{18-2 r}\left(-\frac{1}{3}\right)^1 x^{-r}\right\} \\ & =\left(1+x+2 x^2\right) \\ & \quad\left\{\sum_{r=0}^9{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3r}\right\} \end{aligned} \end{aligned} $

$ \begin{aligned} =\sum_{r=0}^9{ }^9 C_r & \left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r} \\ & +\sum_{r=0}^9{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{19-3 r} \\ & +2\left\{\sum_{r=0}^9{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{20-3 r}\right\} \end{aligned} $

Clearly, first term independent of $x$ by

$ 18-3 r=0 \Rightarrow r=6 $

$\therefore$ Coefficient of independent of $x$ is

$ \begin{aligned} & { }^9 C_6\left(\frac{3}{2}\right)^3\left(-\frac{1}{3}\right)^6={ }^9 C_3 \times \frac{3^3}{8} \times \frac{1}{3^6} \\ & =\frac{9 \times 8 \times 7}{3 \times 2} \times \frac{1}{8} \times \frac{1}{3^3}=\frac{7}{18} . \end{aligned} $

To find the coefficient of $x$ in the expansion of

$ \frac{(1-4x)^2\left(1-2x^2\right)^{\frac{1}{2}}}{(4-x)^{\frac{3}{2}}} $

we proceed by approximating each component in the product.

Expansion of $(1-4x)^2$:

$ (1-4x)^2 = 1 - 8x + 16x^2 $

Approximation of $\left(1-2x^2\right)^{\frac{1}{2}}$:

Using the binomial approximation for small $x^2$, we have:

$ \left(1-2x^2\right)^{\frac{1}{2}} \approx 1 - x^2 $

Approximation of $\left(1-\frac{x}{4}\right)^{-\frac{3}{2}}$:

Again, using binomial expansion, we write:

$ \left(1-\frac{x}{4}\right)^{-\frac{3}{2}} \approx 1 + \frac{3}{2} \times \frac{x}{4} = 1 + \frac{3x}{8} $

Putting it all together, we compute the product:

$ \frac{1}{8} \left[(1-8x+16x^2)(1-x^2)(1+\frac{3x}{8})\right] $

Next, we multiply these expressions and collect the terms involving $x$:

$ \begin{align*} &= \frac{1}{8} \left[(1-8x+16x^2)(1-x^2)(1+\frac{3x}{8})\right] \\ &= \frac{1}{8} \left[(1-8x+16x^2)(1-x^2 + \frac{3x}{8} - \frac{3x^3}{8} + \cdots)\right] \end{align*} $

We are interested only in terms involving $x$. Consider:

The coefficient of $x$ from $(1-8x+16x^2)(1)$ is $-8$.

The contribution due to combining $1$ from $(1-8x+16x^2)$ with the $\frac{3x}{8}$ term from $(1+\frac{3x}{8})$ is $\frac{3}{8}$.

Thus, the total coefficient of $x$ from this product is:

$ \frac{1}{8} \left(\frac{3}{8} - 8\right) $

Simplifying, we calculate:

$ \frac{1}{8} \cdot \left(-\frac{61}{8}\right) = -\frac{61}{64} $

Therefore, the coefficient of $x$ is $-\frac{61}{64}$.

Given that

$P$ is the greatest divisor of $49^n+16 n-1$, for $\forall$ all $n \in N$.

$ \begin{aligned} & 49^n+16 n-1=(1+48)^n+16 n-1 \\ = & 1+n \cdot 48+\frac{n(n-1)}{2} \cdot(48)^2+\ldots . .16 n-1 \\ = & 64 n+{ }^n C_2(48)^2+{ }^n C_3(48)^3+\ldots(48)^n \\ = & 64\left(n+{ }^n C_2 6^2+{ }^n C_3 6^3 \cdot 8+\ldots 6^n \cdot 8^{n-2}\right) \end{aligned} $

So, clearly it is divisible by 64 .

Now, number factor of 64 is

$ 64=2^6=(6+1)=7 $

Given that coefficient of $r$ th: $(r+1)$ th : $(r+2)$ th $=4: 15: 42$ in expansion of $(1+x)^n$.

$ \begin{aligned} & \Rightarrow{ }^n C_{r-1}:{ }^n C_r:{ }^n C_{r+1}=4: 15: 42=K \\ & \Rightarrow \frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{4 K}{15 K} \Rightarrow \frac{r}{n-r+1}=\frac{4}{15} \,\,\,\,\,\,\,\,\,\,\,\,\cdots (i)\\ & \text { and } \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{15 K}{42 K} \Rightarrow n-r=\frac{14(r+1)}{5} \,\,\,\,\,\,\,\,\,\,\,\,\ldots(ii) \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & \frac{5 r}{14 r+19}=\frac{4}{15} \\ \Rightarrow & 75 r=56 r+76 \Rightarrow 19 r=76 \\ \Rightarrow & r=4 \\ \Rightarrow & n-r=\frac{14 \times(4+1)}{5}=14 \text { [from Eq. (i) } \end{aligned} $

Given that coefficients of $(2 r+6)$ th and $(r-1)$ th terms in the expansion of $(1+x)^{21}$ are equal.

$ \begin{array}{rlrl} T_{2 r+5+1} & ={ }^{21} C_{2 r+5}(1)^{21-2 r-5} x^{2 r+5} \\ T_{r-2+1} & ={ }^{21} C_{r-2}(l)^{21-r+2} x^{r-2} \\ \because \quad{ }^{21} C_{2 r+5} & ={ }^{21} C_{r-2} \\ \Rightarrow \quad 2 r+5 & =r-2 \\ r & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(not \,\,possible)\end{array} $

So, $\quad 2 r+5=21-(r-2)$

$ \begin{array}{ccl} & &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, {\left[\because{ }^n C_r={ }^n C_{n-r}\right]} \\ \Rightarrow & & 2 r+5=21-r+2 \\ \Rightarrow & & 3 r=18 \\ \Rightarrow && r=6 \end{array} $

We know that $r$ th term in the expansion $(x+a)^n$ is given by

$ T_{r+1}={ }^n C_r x^{n-r} a^r $

$ \begin{aligned} given,\,\,\,& T_2={ }^n C_1 x^{n-1} a=96 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & T_3={ }^n C_2 x^{n-2} a^2=216 \,\,\,\,\,\,\,\,\,...(ii)\\ & T_4={ }^n C_3 x^{n-3} a^3=216 \,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On dividing Eq. (ii) from Eq. (i), we get

$ \frac{{ }^n C_2 x^{n-2} a^2}{{ }^n C_1 x^{n-1} a}=\frac{216}{96} $

$ \begin{aligned} & \frac{n!}{(n-2)!2!} x^{n-1-1} a^2 \\ & \frac{n!}{(n-1)!1!} x^{n-1} a=\frac{9}{4} \\ & \frac{1}{2} \cdot \frac{(n-1)(n-2)!}{(n-2)!} \frac{a}{x}=\frac{9}{4} \\ &(n-1) \frac{a}{x}=\frac{9}{2} \\ &(n-1) a=\frac{9}{2} x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{aligned} $

Similarly, on dividing Eq. (iii) from Eq. (ii), we get

$ \begin{aligned} \frac{{ }^n C_3 x^{n-3} a^3}{{ }^n C_2 x^{n-2} a^2} & =\frac{216}{216} \\ \Rightarrow \frac{\frac{n!}{(n-3)!3!} x^{n-2-1} a^3}{\frac{n!}{(n-2)!2!} x^{n-2} a^2} & =1 \\ \Rightarrow \frac{2 \cdot 1(n-2)(n-3)!a}{3 \cdot 2 \cdot l(n-3)!} \frac{a}{x} & =1 \\ \Rightarrow \quad(n-2) a & =3 x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{aligned} $

Now, on dividing Eq. (v) from Eq. (iv) we get,

$ \begin{aligned} \frac{(n-2) a}{(n-1) a}=\frac{3 x}{\frac{9}{2} x} & =\frac{2}{3} \\ \Rightarrow \quad 3 n-6 & =2 n-2 \\ n & =4 \end{aligned} $

From Eq. (v), we get

$ \begin{aligned} (4-2) a & =3 x \\ a & =\frac{3}{2} x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{aligned} $

From Eq (i), we get

$ \begin{aligned} { }^4 C_1 x^{4-1} a^1 & =96 \\ \frac{4!}{3!1!} x^3 a & =96 \\ 4 x^3 \cdot \frac{3}{2} x & =96\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [From Eq. (vi)]\\ x^4=\frac{96}{6} & =16 \\ x & =2 \end{aligned} $

Using Eq. (vi), we get

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, a=3 $

$ \begin{aligned}Clearly,\,\,\,\, & x+a=2+3=4+1 \\ & x+a=n+1 \end{aligned} $

Given the expression:

$ \frac{1}{2}\left(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + \ldots + \infty\right)^{-25} $

First, consider the series inside the parentheses:

$ 1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + \ldots + \infty $

This can be rewritten as:

$ 2 \left(1 + 3x + 6x^2 + \ldots + \infty \right) $

Recognizing the series as a result of differentiating and manipulating a geometric series, it sums to:

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

So,

$ \frac{1}{2} \times 2 \times \frac{1}{(1-x)^3} = \frac{1}{(1-x)^3} $

Thus, the expression evaluates to:

$ \frac{1}{2}\left(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + \ldots + \infty\right)^{-25} = \left(\frac{1}{(1-x)^3}\right)^{-25} = (1-x)^{75} $

Therefore, the number of terms in the expansion of $(1-x)^{75}$ is:

$ 75 + 1 = 76 $

In the expansion of $(1+x)^{12}$, we need to determine the sum of all its terms given the ratio of terms equidistant from the middle is $\frac{1}{256}$.

Let's find the middle term and the equidistant terms:

The $k$-th term is denoted by:

$ T_k = \binom{12}{k} x^k $

The equidistant terms from the middle are:

$ T_{12-k} = \binom{12}{12-k} x^{12-k} $

Since, the ratio of equidistant terms is given by:

$ \frac{T_k}{T_{12-k}} = \frac{\binom{12}{k} x^k}{\binom{12}{12-k} x^{12-k}} = \frac{1}{256} $

Since $\binom{12}{12-k} = \binom{12}{k}$, the equation simplifies to:

$ \frac{x^k}{x^{12-k}} = x^{-12 + 2k} = \frac{1}{256} = 2^{-8} $

This gives us:

$ -12 + 2k = -8 $

Solving for $k$, we find:

$ 2k = 4 \implies k = 2 $

Thus, the equidistant terms are $T_2$ and $T_{10}$.

Now, calculate the sum of all terms in the expansion:

When $x = 2$:

$ S = (1 + 2)^{12} = 3^{12} $

When $x = 4$:

$ S = (1 + 4)^{12} = 5^{12} $

Therefore, the sum of all terms could either be $3^{12}$ or $5^{12}$ based on the value of $x$.

From the given condition, one $g^6$

$ \begin{aligned} & \left({ }^{15} C_{10} x^2 a^{10}\right)^2=\left({ }^{15} C_7 x^8 a^7\right)\left({ }^{15} C_{11} x^4 a^{11}\right) \\ & \Rightarrow \frac{a}{x}=\sqrt{\frac{75}{77}} \end{aligned} $

Now, $(x+a)^{15}=x^{15}\left(1+\frac{a}{x}\right)^{15}$

Consider the expansion of $\left(1+\frac{a}{x}\right)^{15}$

$ \begin{aligned} & T_r=T_{r+1} \Rightarrow \frac{r}{(n-r+1)\left(\frac{a}{x}\right)}=1 \\ & \Rightarrow r=16 \sqrt{\frac{75}{77}}-r \sqrt{\frac{75}{77}} \Rightarrow r=\frac{16 \sqrt{\frac{75}{77}}}{1+\sqrt{\frac{75}{77}}} \\ & r=7 \end{aligned} $

If $r=7 \Rightarrow T_7 < T_8$ and $r=8 \Rightarrow T_8>T_9$ Hence, $T_8$ is the greatest term.

$(r+1)$ th term in the given expansion is given by

$ t_{r+1}={ }^{10} C_r(2)^{\frac{10-r}{2}} \cdot 3^{\frac{r}{5}} $

where $r=0,1,2, \ldots 10$

For rational terms

$r=a$ multiple of $5=0,5,10$

$10-r=a$ multiple of $2=0,2,4$

From Eqs. (i) and (ii) possible values of are 0 and 10

$\therefore$ Sum of rational terms

$ \begin{aligned} & =t_1+t_{11}={ }^{10} C_0(\sqrt{2})^{10} \\ & \left(3^{\frac{1}{5}}\right)^0+{ }^{10} C_{10}(\sqrt{2})^0\left(3^{\frac{1}{5}}\right)^{10} \\ & =2^5+3^2=32+9=41 \end{aligned} $

Coefficient of $x^5$ in $\left(a+\frac{x}{5}\right)^{65}$ is

$ { }^{65} C_5 \cdot a^{60} \cdot\left(\frac{1}{5}\right)^5 $

and coefficient of $x^6$ in $\left(a+\frac{x}{5}\right)^{65}$ is

$ { }^{65} C_6 a^{59}\left(\frac{1}{5}\right)^6 $

Now, coefficient of $x^5=$ Coefficient of $x^6$ in expansion $\left(a+\frac{x}{5}\right)^{65}$

$ \begin{aligned} & { }^{65} C_5 \cdot a^{60} \cdot\left(\frac{1}{5}\right)^5={ }^{65} C_6 \cdot a^{59}\left(\frac{1}{5}\right)^6 \\ \Rightarrow & \frac{65!}{5!60!} \cdot a^{60} \cdot \frac{1}{5^5}=\frac{65!}{6!59!} a^{59} \cdot \frac{1}{5^6} \\ \Rightarrow & \frac{a}{60}=\frac{1}{6} \times \frac{1}{5} \Rightarrow a=\frac{60}{30}=2 \\ \Rightarrow & a=2 \end{aligned} $

$\therefore$ Coefficient of $x^2$ in expansion of $\left(2+\frac{x}{5}\right)^4$ is

$ \begin{aligned} & ={ }^4 C_2 \cdot(2)^2 \cdot\left(\frac{1}{5}\right)^2 \\ & =\frac{4!}{2!2!} \cdot \frac{4}{25}=\frac{24}{25} \end{aligned} $

Given the expression $(3x-2)^{\frac{2}{3}}$, we can rewrite it as:

$ (3x - 2)^{\frac{2}{3}} = (-2 + 3x)^{\frac{2}{3}} $

This can be further expressed as:

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

To find the 4th term of the expansion of $2^{\frac{2}{3}}\left(-1 + \frac{3}{2}x\right)^{\frac{2}{3}}$, we use the formula for the binomial expansion term, which is given by:

$ T_4 = 2^{\frac{2}{3}} \cdot \binom{\frac{2}{3}}{3} \cdot (-1)^{\frac{2}{3} - 3} \cdot \left(\frac{3}{2}x\right)^3 $

Calculating the binomial coefficient:

$ \binom{\frac{2}{3}}{3} = \frac{\frac{2}{3} \left(\frac{2}{3} - 1\right) \left(\frac{2}{3} - 2\right)}{3 \times 2 \times 1} $

$ = \frac{\frac{2}{3} \cdot \left(-\frac{1}{3}\right) \cdot \left(-\frac{4}{3}\right)}{6} $

Now, simplifying:

$ = \frac{2^{\frac{2}{3}} \cdot \left(-\frac{2}{27}\right)}{6} \cdot (-1)^{\frac{-7}{3}} \cdot \left(\frac{27}{8}\right)x^3 $

$ = -2^{\frac{2}{3}} \cdot \frac{8}{27} \times \frac{27}{8} \times \frac{1}{6} x^3 $

Finally, simplifying gives:

$ = \frac{-\sqrt[3]{4}}{6} x^3 $

So, the 4th term in the expansion is $-\frac{\sqrt[3]{4}}{6} x^3$.

We have, to find coefficient of $x^3$ if the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$.

The function is of the type $\left(a x^p+\frac{b}{x^q}\right)^n$, then

$ r=\frac{n p-k}{p+q}=\frac{5 \times 3-5}{3+2}=\frac{5(3-1)}{5}=2 $

Coefficient of $T_{r+1}={ }^n c_r a^{n-r} \cdot b^t$

Coefficient of $T_3=5 C_2 \cdot(2)^{5-2} \cdot\left(\frac{1}{3}\right)^2$

$ \begin{aligned} & =\frac{51}{3!\cdot 2!} \times 2^3 \times \frac{1}{9} \\ & =\frac{5 \times 4}{2} \times 2^3 \times \frac{1}{9}=\frac{80}{9} \end{aligned} $

To find the numerically greatest term in the expansion of $(5+3x)^6$ when $x=1$, we start by analyzing the general term in the binomial expansion:

$ T_{r+1} = {^nC_r} \cdot x^{n-r} \cdot a^r $

For our specific case:

$ T_{r+1} = {^6C_r} \cdot 5^{6-r} \cdot (3x)^r $

We are looking for $\frac{T_{r+1}}{T_r}$ to determine when the terms start decreasing:

$ \begin{aligned} \frac{T_{r+1}}{T_r} &= \left(\frac{6-r+1}{r}\right) \cdot \frac{3x}{5} \\ &= \frac{21-3r}{5r} \quad \text{(by substituting $x=1$)} \end{aligned} $

Now, we analyze when $T_{r+1} > T_r$:

$ \begin{aligned} 21 - 3r &> 5r \\ 8r &< 21 \\ r &< 2.625 \end{aligned} $

Thus, to find the numerically largest term, we evaluate $r$ when $T_{r+1} > T_r$ and $T_{r+1} < T_r$. Specifically, when:

$ \begin{aligned} r < 2.625 & : T_{r+1} > T_r \\ r > \frac{21}{8} & : T_{r+1} < T_r \end{aligned} $

Now let's calculate $T_3$ and $T_4$:

For $T_3$:

$ \begin{aligned} T_3 &= {^6C_2} \cdot 5^4 \cdot (3)^2 \cdot 1^2 \\ &= 15 \cdot 625 \cdot 9 \\ &= 5^5 \cdot 3^3 \end{aligned} $

For $T_4$:

$ \begin{aligned} T_4 &= {^6C_3} \cdot 5^3 \cdot (3)^3 \cdot 1^3 \\ &= 20 \cdot 125 \cdot 27 \\ &= 4 \times 5^3 \times 3^3 \end{aligned} $

The calculations show that the largest term is the 3rd term ($T_3$), which equals $5^5 \times 3^3$. Thus, the numerically greatest term in the expansion is $5^5 \times 3^3$.

$ \begin{aligned} & T_{r+1}={ }^{36} C_r\left(\frac{2 x^2}{5}\right)^{10-r} \times\left(\sqrt{\frac{5}{x}}\right)^r \\ &{ }^{10} C_r \frac{(2)^{32-}}{5^{10-r-\frac{1}{2}}} \times x^{30-3-\frac{r}{2}} \\ &={ }^{10} C_r \frac{(2)^{10-1}}{5^{31-\frac{y}{2}}} \times x^{30-\frac{v}{2}} \end{aligned} $

Term will be independent if

$ 20-\frac{5 r}{2}=0 \Rightarrow r=8 $

Independent term

$ \begin{aligned} & ={ }^{10} C_3 \frac{(2)^{10-1}}{5} \\ & ={ }^{10} C_2 \times 2^2 \times 5^3=4500 \end{aligned} $

Square root of $4500=30 \sqrt{5}$

We have, $\left(3+x+x^2\right)^6$

General term of the expansion

$ =\frac{6!}{r!s!t!} \times 3^r \times x^5 \times\left(x^2\right)^t $

Where,

$ \begin{aligned} & r+s+t=6 \\ & =\frac{6!}{r!s!t!} \times 3^r \times x^{5+2 t} \end{aligned} $

For the coefficient of $x^5$,

$ s+2 t=5 $

But $r+s+t=6$

$ \begin{aligned} & r+5-t=6 \\ & r-a=1+t \text { and } s=5-2 t \end{aligned} $

Now, $t=0$, then $r=1, s=5$

$ \begin{aligned} & t=1, \text { then } r=2, s=3 \\ & t=2 \text { then } r=3, s=1 \end{aligned} $

$\therefore$ There are only 3 terms that contain $x^5$

$\therefore$ Coefficient of $x^5$

$ \begin{aligned} & =\frac{6!}{0!1!5!} \times 3^1+\frac{6!}{1!2!3!} \times 3^2+\frac{6!}{2!3!!!} \times 3^3 \\ & =18+540+1620=2178 \end{aligned} $

To determine the absolute value of the difference between the coefficients of $x^4$ and $x^6$ in the expansion of the given expression, we start with:

$ x^2 - 2x^2 + (x + 1)^4(x^2 - 1)^2 $

We simplify the function as follows:

$ f(x) = \frac{2x^2}{(x^2+1)(x^2+2)} $

Expanding, we have:

$ f(x) = 2x^2(x^2+1)^{-1}(x^2+2)^{-1} $

Simplifying further gives:

$ f(x) = \frac{2x^2}{2} \left(1 + x^2\right)^{-1}\left(1 + \frac{x^2}{2}\right)^{-1} $

This can be expanded into:

$ f(x) = x^2[1 - x^2 + x^4 - x^6 + \cdots] \cdot \left[1 - \frac{x^2}{2} + \frac{x^4}{4} - \frac{x^4}{8} + \cdots\right] $

Now, let’s find the coefficients.

For $x^4$:

Coefficient calculation:

$\left[-\frac{1}{2} - 1\right] = \frac{-3}{2}$

For $x^6$:

Coefficient calculation:

$\left[\frac{1}{4} + \frac{1}{2} + 1\right] - \frac{7}{4}$

Now calculate the absolute difference:

$ \text{Difference} = \left| \frac{7}{4} - \left(\frac{-3}{2}\right) \right| = \left| \frac{7}{4} + \frac{3}{2} \right| $

$ = \left| \frac{7}{4} + \frac{6}{4} \right| = \left| \frac{13}{4} \right| = \frac{13}{4} $

This process allows us to correctly determine the absolute value of the difference between the coefficients of $x^4$ and $x^6$ in the expansion.

Given the inequality $\binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3}$:

We start with the given expression:

$ \binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3} $

Using the properties of combinations, we know that:

$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} $

Applying this, we get:

$ \binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3} $

This simplifies to:

$ \frac{(n-1)!}{3!(n-4)!} + \frac{(n-1)!}{4!(n-5)!} > \frac{n!}{3!(n-3)!} $

Rewriting and dividing both sides by common terms:

$ \frac{1}{4!} + \frac{1}{(n-4)} > \frac{1}{{n-3 \cdot n-4}} $

Simplifying further:

$ \frac{1}{4} > \frac{1}{(n-3)(n-4)} $

Taking the reciprocal, we have:

$ 4 < (n-3)(n-4) $

To find the minimum value of $n$:

$ (n-3)(n-4) > 4 $

Solving this inequality, we find:

$ n-3 > 4 $

Thus:

$ n > 7 $

The smallest integer $n$ satisfying the inequality is $n = 8$.

Hence, the least value of $n$ is 8.