Binomial Theorem

$(x-2 y+3 z)^6$

By multinomial theorem,

$ \begin{aligned} & \Sigma \frac{6!}{p!q!r!}(x)^p(-2 y)^q(3 z)^r \\\\ \Rightarrow \quad & \Sigma \frac{6!}{p!q!r!}(-2)^q(3)^4 x^p y^f z^r \end{aligned} $

Then, coefficient of $x y^2 z^3$ is

$ \begin{aligned} & p=1, q=2 r=3 \\\\ \Rightarrow \quad & \frac{6!}{1!2!3!}(-2)^2(3)^3 \\\\ & =\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 2} \times 4 \times 27 \\\\ & =240 \times 27=6480 \end{aligned} $

$\left(125 x^2-\frac{27}{x}\right)^{-\frac{2}{3}}$

$\Rightarrow \quad 5^{-2}\left(x^2-\frac{27}{125 x}\right)^{-\frac{2}{3}}$

$\Rightarrow\left|\frac{27}{125 x^3}\right| \leq 1 \Rightarrow-1 \leq \frac{27}{125 x^3} \leq 1$

$\Rightarrow \quad-1 \leq\left(\frac{3}{5 x}\right)^3 \leq 1 \Rightarrow-1 \leq \frac{3}{5 x}<1$

$\Rightarrow \quad-\frac{5}{3}<\frac{1}{x}<\frac{5}{3} \Rightarrow \frac{1}{x}<\frac{5}{3}$

$ \begin{array}{l}\Rightarrow \quad \frac{3}{5} < x < \infty \\\\ \text { and } \quad \frac{1}{x} > -\frac{5}{3} \\\\ \Rightarrow \quad-\infty < x < -\frac{3}{5} \\\\ \therefore \quad x \in\left(-\infty,-\frac{3}{5}\right) \cup\left(\frac{3}{5}, \infty\right)\end{array} $

Given the expression $3^{2n+2} - 8n - 9$, we want to determine the maximum power of $2$ that divides this expression for all $n \in \mathbb{N}$.

First, we can rewrite $3^{2n+2}$ as $9^{n+1}$:

$ 9^{n+1} - 8n - 9 $

Considering $9 = 1 + 8$, we can apply the binomial expansion to $ (1 + 8)^{n+1} $:

$ (1 + 8)^{n+1} = {^{n+1}C_0} + {^{n+1}C_1} \cdot 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots $

Simplifying, we get:

$ 1 + 8(n+1) + 8^2 {^{n+1}C_2} + {^{n+1}C_3} \cdot 8^3 + \ldots $

Subtracting $8n + 9$ from this sequence, we have:

$ 1 + 8n + 8 + 8^2 {^{n+1}C_2} + \ldots - 8n - 9 $

Which simplifies to:

$ 8^2 ({^{n+1}C_2} + 8 {^{n+1}C_3} + \ldots) $

The expression $8^2$ can be further evaluated as $2^6$. Thus, the binomial expansion shows that after simplification, the given expression is a multiple of $2^6$.

Therefore, the maximum value of $p$ such that the expression is divisible by $2^p$ is:

$ \boxed{6} $

Given the polynomial expression $(1 + x + x^2 + x^3)^{100}$, we are interested in the coefficient of $x^r$, denoted by $a_r$. To find the sum of the products $r \cdot a_r$, we perform the following steps:

First, evaluate the polynomial at $x = 1$:

$ (1 + 1 + 1^2 + 1^3)^{100} = (4)^{100} $

This shows that the sum of all coefficients, $\sum_{r=0}^{300} a_r$, equals $4^{100}$.

Next, differentiate $(1 + x + x^2 + x^3)^{100}$ with respect to $x$:

$ \frac{d}{dx}\left((1 + x + x^2 + x^3)^{100}\right) = 100(1 + x + x^2 + x^3)^{99}(1 + 2x + 3x^2) $

Substitute $x = 1$ into the derivative:

$ \begin{aligned} \sum_{r=0}^{300} r \cdot a_r &= 100(4)^{99}(1 + 2 \cdot 1 + 3 \cdot 1^2) \\ &= 100(4^{99}) \cdot 6 \end{aligned} $

Simplify the expression to find:

$ \sum_{r=0}^{300} r \cdot a_r = 600(4^{99}) = 150 \times 4^{100} $

Since $4^{100} = \sum_{r=0}^{300} a_r = S$, we have:

$ \sum_{r=0}^{300} r \cdot a_r = 150S $

Thus, $\sum_{r=0}^{300} r \cdot a_r$ is equal to $150S$.

Given the binomial distribution $ X \sim B(6, p) $ and the condition $ \frac{P(X=4)}{P(X=2)} = \frac{1}{9} $, let's solve for $ p $.

Start with the equation:

$ 9 P(X=4) = P(X=2) $

For a binomial distribution:

$ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} $

Using this formula, we set up the probabilities:

$ P(X=4) = \binom{6}{4} p^4 (1-p)^2 $

$ P(X=2) = \binom{6}{2} p^2 (1-p)^4 $

We substitute these into our equation:

$ 9 \times \binom{6}{4} p^4 (1-p)^2 = \binom{6}{2} p^2 (1-p)^4 $

Calculate the binomial coefficients:

$ \binom{6}{4} = 15 $

$ \binom{6}{2} = 15 $

Substitute these values:

$ 9 \times 15 p^4 (1-p)^2 = 15 p^2 (1-p)^4 $

Cancel out the common term $ 15 $:

$ 9 p^4 (1-p)^2 = p^2 (1-p)^4 $

Divide both sides by $ p^2 (1-p)^2 $ (assuming $ p \neq 0 $ and $ p \neq 1 $):

$ 9 p^2 = (1-p)^2 $

Take the square root of both sides:

$ 3 p = 1-p $

Solve for $ p $:

$ 3p + p = 1 $

$ 4p = 1 $

$ p = \frac{1}{4} $

Therefore, the value of $ p $ is $\frac{1}{4}$.

We know that $ T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r} $

$ \begin{aligned} T_{6+1} & ={ }^{8} C_{6}\left(\frac{5}{p^{3}}\right)^{8-6}\left(-\frac{3 q}{7}\right)^{6}=700 \\\\ & =28 \times \frac{25}{p^{6}} \times \frac{729 q^{6}}{7^{6}}=700 \\\\ \Rightarrow(7 p)^{6} & =\frac{28 \times 25 \times 729}{700} q^{6} \\\\ \Rightarrow(7 p)^{6} & =(3 q)^{6} \Rightarrow 7 p= \pm 3 q \Rightarrow 49 p^{2}=9 q^{2} \end{aligned} $

Given,

$ \left(5 x+\frac{7}{x}\right)^{-3 / 2}=(5 x)^{-\frac{3}{2}}\left(1+\frac{7}{5 x^{2}}\right)^{-3 / 2} $

We know that

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

$ \begin{array}{l} \therefore T_{4}=\frac{1}{5 x \sqrt{5 x}}\left[\frac{-3}{2} \times \frac{5}{2} \times \frac{7}{2} \times \frac{1}{6} \times\left(\frac{7}{5 x^{2}}\right)^{3}\right] \\\\ \begin{aligned} \left(x^{7} \sqrt{5 x}\right) T_{4} & =-\frac{1}{5} \times \frac{3}{2} \times \frac{5}{2} \times \frac{7}{2} \times \frac{1}{6} \times \frac{7^{3}}{5^{3}} \\\\ & =-\frac{7^{4}}{2^{4} 5^{3}} \end{aligned} \end{array} $

Let 3 consecutive terms in the expansion of $(1+x)^{23}$ be

$ T_{r+1}, T_{r+2} \text { and } T_{r+3} $

Coefficient of $T_{r+1}={ }^{23} C_r$

Coefficient of $T_{r+2}={ }^{23} C_{r+1}$

Coefficient of $T_{r+3}={ }^{23} C_{r+2}$

According to the question,

$ \begin{array}{ll} { }^{23} C_{r+1}-{ }^{23} C_r & ={ }^{23} C_{r+2}-{ }^{23} C_{r+1} \\\\ \Rightarrow \quad & \frac{23!}{(r+1)!(22-r)!}-\frac{23!}{r!(23-r)!} \\\\ & =\frac{23!}{(r+2)!(21-r)!}-\frac{23!}{(r+1)!(22-r)!} \\\\ \Rightarrow \quad & \frac{23-r}{(r+1)!(23-r)!}-\frac{(r+1)}{(r+1)!(23-r)!} \\\\ & =\frac{22-r}{(r+2)!(22-r)!}-\frac{r+2}{(r+2)!(22-r)!} \\\\ \Rightarrow \quad & (r+2)[(23-r)-(r+1)] \\\\ \Rightarrow \quad & =(23-r)[(22-r)-(r+2)] \\\\ \Rightarrow \quad & (r+2)(22-2 r)=(23-r)(20-2 r) \\\\ \Rightarrow \quad & r^2-84 r+416=0 \\\\ \Rightarrow \quad & (r-8)(r-13)=0 \\\\ \Rightarrow \quad r=8,13 \end{array} $

Hence, 3 consecutive terms are either $T_9, T_{10}$ and $T_{11}$ or $T_{14}, T_{15}$ and $T_{16}$.

Let the numerically greatest term in the expansion of $(3 x-16 y)^{15}$ when $x=\frac{2}{3}$ and $y=\frac{3}{2}$ is $T_{r+1}$

$ \begin{aligned} T_{r+1} & ={ }^{15} C_r\left(3\left(\frac{2}{3}\right)\right)^{15-r}\left(-16\left(\frac{3}{2}\right)\right)^r \\\\ & ={ }^{15} C_r(2)^{15-r}(-24)^r \end{aligned} $

Clearly, the value the expression will be maximum only when the value $r$ is largest possible even integer.

$ \text { So, } r=14 \quad[\because 0 \leq r \leq 15 \text { and } r \in Z] $

$\therefore 15$ th term will be the numerically greatest term.

To find largest positive integer $K$, that divides $81^n+20 n-1$ for $n \in N$, we start evaluating the expression for small values of $n$

For $n=1: 81^1+20.1-1=100$

For $n=2: 81^2+20.2-1=6600$

Now,

The greatest common divisor of 100 and 6600 is 100 . Hence, $K=100$ is the largest integer that divides

$81^n+20 n-1$ for all $n \in N$

and the divisior of 100 are

$ \begin{aligned} & 1,2,4,5,10,20,25,50,100 \\\\ & \begin{aligned} \Rightarrow S=1+2+4+5+10+20+25+50+100=217 \end{aligned} \end{aligned} $

Hence, $S-K=117$

$ \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.

General term of given binomial expansion is

$ \begin{aligned} T_{r+1} & ={ }^{10} C_{r}\left(\frac{1}{3} x^{\frac{1}{2}}\right)^{10-r}\left(x^{-\frac{1}{4}}\right)^{r} \\ & ={ }^{10} C_{r} \times\left(\frac{1}{3}\right)^{10-r} x^{\frac{10-r}{2}-\frac{r}{4}} \end{aligned} $

We have to find coefficient of $ x^{2} $.

$ \begin{aligned} \frac{10-r}{2}-\frac{r}{4} & =2 \Rightarrow r=4 \\ \Rightarrow \quad T_{4+1} & ={ }^{10} C_{4}\left(\frac{1}{3}\right)^{6} x^{2} \\ \therefore \text { Coefficient of } x^{2} & ={ }^{10} C_{4}\left(\frac{1}{3}\right)^{6} \\ & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{1}{3^{6}}=\frac{70}{243} \end{aligned} $

We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.

We need to find the value of $2(a+b+c)$.

Now we are given $p_1 = 20$ and $p_2 = 210$. So, $ 10a^9 b = 20 \implies a^9 b = 2 $

and $ 45a^8 b^2 + 10a^9 c = 210 $

Now, divide the second equation by $a^8$: $ 45b^2 + 10ac = 210 $

We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides: $ ab = \sqrt[9]{2} $

Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as: $ 45k^2 + 10k^9 = 210 $

From the equation $ab = k = \sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:

$ a+b = 2 $

and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:

$ 45a^8 b^2 + 10a^9 c = 210 $

Using $a=1$ and $b=2$, we get:

$ 45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3 $

So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:

$ 2(a+b+c) = 2(1+2+3) = 2(6) = 12 $

Therefore, the answer is $\boxed{12}$.

If ${I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}$

${I_1} + f - f'=$ Even

${I_1} = $ Even

${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$

= Even

${I_2} = $ Even

For $\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$

${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$

$\therefore$ ${x^{15}} \to 3(15 - r) - {r \over 3} = 15$

$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$

Similarly, for ${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$

${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$

$\therefore$ For ${x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6$

$\therefore$ ${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$