Binomial Theorem

We have, $x^{3 / 2}(3+x)^{1 / 2}$

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

Now, coefficient of general term of $\left(1+\frac{3}{x}\right)^{1 / 2}$ is

$ \begin{aligned} & \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{1}{3}\right) \ldots\left(\frac{1}{2}-k+1\right)}{k!} \\ & \quad \frac{(-1)^{k+1} 1 \cdot 3 \cdot 5 \ldots(2 k-3)}{2 k k!} \end{aligned} $

On multiply by $x^2$, then

$ x^2\left(1+\frac{3}{x}\right)^{1 / 2}=\sum_{k=0}^{\infty}{ }^{1 / 2} C_k 3^k x^{2-k} $

For coefficient of $\frac{1}{x^n}, 2-k=-n$

$ \Rightarrow \quad k=n+2 $

Coefficient of $\frac{1}{x^n}$ is

$ \begin{aligned} & \frac{(-1)^{n+3} 1 \cdot 3 \cdot 5 \ldots(2(n+2)-3) 3^{n+2}}{2^{n+2}(n+2)!} \\ & \quad=\frac{(-1)^{n+1} 1 \cdot 3 \cdot 5 \ldots(2 n+1)}{2^{n+1}(n+2)!} 3^{n+2} \end{aligned} $

Given, coefficient of 3rd term from beginning in the expansion $\left(a x^2-\frac{8}{b x}\right)^9 =$ Coefficient of 3rd term from end in the expansion $\left(a x-\frac{2}{b x^2}\right)^9$

For $\left(a x^2-\frac{8}{b x}\right)^9$

$ \begin{aligned} T_3 & ={ }^9 C_2\left(a x^2\right)^7\left(\frac{-8}{b x}\right)^2 \\ & =\frac{9!}{2!7!} a^7 x^{14} \cdot \frac{64}{b^2 x^2} \\ & =\frac{36 \cdot a^7}{b^2} \cdot x^{12} \cdot 64 \end{aligned} $

So, its coefficient is $\frac{36 \cdot 64 a^7}{b^2}$

For $\left(a x-\frac{2}{b x^2}\right)^9$

$ \begin{aligned} T_8 & ={ }^9 C_7(a x)^2\left(\frac{-2}{b x^2}\right)^7 \\ & =\frac{9!}{2!7!} a^2 x^2 \cdot \frac{-128}{b^7 x^{14}} \end{aligned} $

So, if coefficient is

$ =-36 \times 128 \frac{a^2}{b^7} $

Now, $36 \cdot \frac{64 a^7}{b^2}=-36 \cdot 128 \frac{a^2}{b^7}$

$ \begin{aligned} \Rightarrow & & 64 \cdot a^5 b^5 & =-128 \\ \Rightarrow & & a^5 b^5 & =-2 \end{aligned} $

Given, $5^{2 n}-48 n+k$ is divisible by 24

Let $\quad E(n)=5^{2 n}-48 n+k$

and $\quad E(n) \equiv 0(\bmod 24), \forall n \in N$

$\Rightarrow \quad 5^{2 n}-48 n+k \equiv 0(\bmod 24)$

$\Rightarrow \quad k=48 n-5^{2 n}(\bmod 24)$

For $n=1$,

$ \begin{aligned} k & =48(1)-5^{2 \times 1}(\bmod 24) \\ & =48-25(\bmod 24) \\ & =23(\bmod 24) \end{aligned} $

For $n=2$

$k=48(2)-5^1(\bmod 24)$

$ \begin{aligned} & =96-625(\bmod 24) \\ & =529(\bmod 24)=23(\bmod 24) \end{aligned} $

Thus, divisibility for 24 to hold for all $n$, we must have $k \equiv 23(\bmod 24)$

Thus, the least positive integral value of $k$ is 23 .

Given $X \sim B(7, P)$ is a binomial variate and $P(X=3)=P(X=5)$

Using the binomial probability formula

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

Here $n=7$,

So, $P(X=3)=\binom{7}{3} P^3(1-P)^4$

And, $P(X=5)=\binom{7}{5} P^5(1-P)^2$

$ \begin{aligned} & \therefore\binom{7}{3} P^3(1-P)^4=\binom{7}{5} P^5(1-P)^2 \\ & \Rightarrow 35 P^3(1-P)^4=21 P^5(1-P)^2 \\ & \Rightarrow 35(1-P)^2=21 P^2, \text { where } P \neq 0, P \neq 1 \\ & \Rightarrow 5(1-P)^2=3 P^2 \\ & \Rightarrow 5\left(1-2 P+P^2\right)=3 P^2 \\ & \Rightarrow 5-10 P+5 P^2-3 P^2=0 \\ & \Rightarrow 2 P^2-10 P+5=0 \\ & \Rightarrow P=\frac{-(-10) \pm \sqrt{100-4 \times 2 \times 5}}{2 \times 2} \\ & \quad=\frac{10 \pm \sqrt{100-40}}{4} \\ & \quad=\frac{10 \pm 2 \sqrt{15}}{4}=\frac{5 \pm \sqrt{15}}{2} \end{aligned} $

But $P=\frac{5+\sqrt{15}}{2}>1$, So it can't be a valid probability.

$ \therefore P=\frac{5-\sqrt{15}}{2} $

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} & \text { }{ }^n C_{n-r}+3 \cdot{ }^n C_{n-r+1}+3 \cdot{ }^n C_{n-r+2}+{ }^n C_{n-r+3}={ }^x C_r \\ & \Rightarrow\left({ }^n C_{n-r}+{ }^n C_{n-r+1}\right)+2 \cdot\left({ }^n C_{n-r+1}+{ }^n C_{n-r+2}\right) +\left({ }^n C_{n-r+2}+{ }^n C_{n-r+3}\right)={ }^n C_r \end{aligned} $

$ \begin{aligned} & \Rightarrow{ }^{n+1} C_{n-r+1}+2^{n+1} C_{n-r+2}+{ }^{n+1} C_{n-r+3}={ }^x C_r \\ & {\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right] } \end{aligned} $

$ \begin{aligned} \Rightarrow\left\{{ }^{n+1} C_{n-r+1}+{ }^{n+1} C_{n-r+2}\right\} +\left\{{ }^{n+1} C_{n-r+2}+{ }^{n+1} C_{n-r+3}\right\}={ }^x C_r \end{aligned} $

$ \begin{aligned} & \Rightarrow{ }^{n+2} C_{n-r+2}+{ }^{n+2} C_{n-r+3}={ }^x C_r \\ & \Rightarrow{ }^{n+3} C_{n-r+3}={ }^x C_r \end{aligned} $

$ \begin{aligned} &\Rightarrow{ }^{n+3} C_r={ }^x C_r\,\,\,\,\,\,\left[\because{ }^{n+3} C_{n-r+3}={ }^{n+3} C_r\right] \end{aligned} $

$ \therefore \quad x=n+3 $

We have, $(1+x)^{41}\left(1-x+x^2\right)^{40}$

$ \begin{aligned} & =\left\{(1+x)\left(1-x+x^2\right)\right\}^{40}(1+x) \\ & =\left(1+x^3\right)^{40}(1+x) \\ & =\left(1+x^3\right)^{40}+x\left(1+x^3\right)^{40} \end{aligned} $

∴ Coefficient of $x^{50}$ in $(1+x)^{41}\left(1-x+x^2\right)^{40}$

= Coefficient of $x^{50}$ in $\left(1+x^3\right)^{40}+$ Coefficient of $x^{50}$ in $x\left(1+x^3\right)^{40}$

$=$ Coefficient of $x^{50}$ in $\left(1+x^3\right)^{40}+$ Coefficient of $x^{49}$ in $\left(1+x^3\right)^{40}$

= Coefficient of $x^{50}$ in $\sum_{r=0}^{40}{ }^{40} C_r x^{3 r}+$ Coefficient of $x^{50}$ in $\sum_{r=0}^{40}{ }^{40} C_r x^{3 r}=0+0=0$

$ \begin{aligned} & T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\ & ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\ & ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}} \end{aligned} $

For coefficient of $x^{2 / 3}$

$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\ & \Rightarrow 90-16 r=10 \\ & \Rightarrow r=5 \end{aligned}$

For coefficient of $x^{-2 / 5}$

$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\ & \Rightarrow 90-16 r=-6 \\ & \Rightarrow r=6 \end{aligned}$

Sum of coefficient of $x^{2 / 3}$ & $x^{-2 / 5}$

$\begin{aligned} & ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\ & =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4} \end{aligned}$

$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$

It is a G.P. with first term $=x^2(1+x)^{98}$

and common ratio $=\frac{x}{1+x}$

sum of these term $=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$

$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$

$\begin{aligned} & ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\ & \Rightarrow p=68, q=15 \\ & \Rightarrow p+q=83 \end{aligned}$

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

Independent of $x \Rightarrow 20-2 r-3 r=0$

$r=4$

Independent of $x$ is ${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$

$\begin{gathered} \frac{210}{2 \times 8} a^3=105 \\ \Rightarrow \quad a=2 \\ a^2=4 \end{gathered}$

$\begin{aligned} & \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\ & T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r \end{aligned}$

For constant term $-12+r+r=0$

$\begin{aligned} & \Rightarrow \quad r=6 \\ & \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6 \end{aligned}$

$\begin{aligned} & ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\ & =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\ & =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\ & \therefore \quad \alpha=\frac{693}{25} \\ & 25 \alpha=693 \end{aligned}$

$\begin{aligned} & (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\ & { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\ & { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\ & \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\ & \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \\ & \Rightarrow 30 n^2-450 n+1620=5 n^2 \\ & \Rightarrow \frac{1}{n-5}\left[\frac{n-4-5}{5(n-4)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6(n-5)}\right] \\ & \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}\left[\frac{n-11}{6}\right] \\ & \Rightarrow n^2-21 n+98=0 \\ & n_{\max }=14 \end{aligned}$

$\begin{aligned} & T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\ & ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)} \end{aligned}$

For rational terms,

$\frac{r}{3}$ and $\frac{r}{5}$ must be integer

3 and 5 divide $r \Rightarrow 15$ divides $r \Rightarrow r=0$ and $r=15$

${ }^{15} C_0 5^0 2^3+{ }^{15} C_{15} 5^5 2^{(0)}$

$\begin{aligned} & =8+3125 \\ & =3133 \end{aligned}$

Put $x=1$

$\therefore \mathrm{a}=1$

$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$

Using L' HOPITAL Rule

$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}$

Now, $\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0$

$(\mathrm{D}<0)$

$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$

$2-p, p, 2-\alpha, \alpha$

Binomial coefficients are

$\begin{aligned} & { }^n C_r,{ }^n C_{r+1},{ }^n C_{r+2},{ }^n C_{r+3} \text { respectively } \\ \Rightarrow \quad & { }^n C_r+{ }^n C_{r+1}=2 \\ \Rightarrow \quad & { }^{n+1} C_{r+1}=2 \quad \ldots . .(1) \end{aligned}$

Also, ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+2}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+3}=2$

$\Rightarrow \quad{ }^{n+1} C_{r+3}=2$ $\quad\text{..... (2)}$

From (1) and (2)

$\begin{aligned} & { }^{n+1} C_{r+1}={ }^{n+1} C_{r+3} \\ & \Rightarrow \quad 2 \mathrm{r}+4=\mathrm{n}+1 \\ & \mathrm{n}=2 \mathrm{r}+3 \\ & { }^{2 \mathrm{r}+4} \mathrm{C}_{\mathrm{r}+1}=2 \\ \end{aligned}$

Data Inconsistent

${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$

$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$

$\begin{aligned} & \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\ & \frac{r+1}{n}=k^2-8 \\ & \Rightarrow k^2-8>0 \\ & (k-2 \sqrt{2})(k+2 \sqrt{2})>0 \end{aligned}$

$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$

$\begin{aligned} \therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\ \Rightarrow & \mathrm{k}^2-8 \leq 1 \\ & \mathrm{k}^2-9 \leq 0 \\ & -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)} \end{aligned}$

From equation (I) and (II) we get

$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$

Sum of coefficients in the expansion of $\left(1-3 \mathrm{x}+10 \mathrm{x}^2\right)^{\mathrm{n}}=\mathrm{A}$

then $A=(1-3+10)^n=8^n$ (put $x=1$)

and sum of coefficients in the expansion of

$\begin{aligned} & \left(1+x^2\right)^n=B \\ & \text { then } B=(1+1)^n=2^n \\ & A=B^3 \end{aligned}$