General And Middle Terms

$a_r={}^{10}{C}_{10-r}={}^{10}{C}_r$
$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{{}^{10}{C}_r}{{}^{10}{C}_{r-1}}\right)}^2=\sum _{r=1}^{10}{r}^3{\left(\frac{11-r}{r}\right)}^2=\sum _{r=1}^{10}r(11-r{)}^2$
$=\sum _{r=1}^{10}\left(121r+r^3-22r^2\right)=1210$

$a_r={}^{10}{C}_{10-r}={}^{10}{C}_r$
$?\underset{r=1}{\overset{10}{?}}{r}^3{\left(\frac{{}^{10}{C}_r}{{}^{10}{C}_{r-1}}\right)}^2=\underset{r=1}{\overset{10}{?}}{r}^3{\left(\frac{11-r}{r}\right)}^2=\underset{r=1}{\overset{10}{?}}r(11-r{)}^2$
$=\underset{r=1}{\overset{10}{?}}\left(121r+r^3-22r^2\right)=1210$

The correct answer is 5040.
${\left(\frac{4x}{5}+\frac{5}{2x^2}\right)}^9$
Now, $T_{r+1}={}^9{C}_r\cdot {\left(\frac{4x}{5}\right)}^{9-r}{\left(\frac{5}{2x^2}\right)}^r$
$={}^9{C}_r\cdot {\left(\frac{4}{5}\right)}^{9-r}{\left(\frac{5}{2}\right)}^r\cdot x^{9-3r}$
Coefficient of $x^{-6}$ i.e. $9-3r=-6\Rightarrow r=5$
So, Coefficient of $x^{-6}={}^9{C}_5{\left(\frac{4}{5}\right)}^4\cdot {\left(\frac{5}{2}\right)}^5=5040$

We are asked to find the coefficient of in the expansion of the series:

Step 1: Understanding the Series

The given series can be written as:

This is a sum of terms where each term involves multiplied by , where ranges from 0 to 500.

The general term in the expansion is:

Step 2: Finding the Coefficient of

We need to find the coefficient of in the entire series. For each term , the exponent of in the expanded form of will be . We are interested in terms where the total exponent of equals 301.

The exponent of in each term is:

where is the exponent of in the expansion of . The coefficient of in is:

Step 3: Coefficient of in the Series

Thus, the required coefficient is the sum of the coefficients of in all terms. After simplifying, we get the coefficient of as:

Thus, the correct answer is .

$a_2-a_1=a_3-a_2=\dots ..=a_{2022}-a_{2021}=1$.
$\therefore {\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)+\dots ..+{\tan }^{-1}\left(\frac{a_{2022}-a_{2021}}{1+a_{2021}{a}_{2022}}\right)$
$=\left[\left({\tan }^{-1}{a}_2\right)-{\tan }^{-1}{a}_1\right]+\left[{\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2\right]+\dots ..+\left[{\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_{2021}\right]$
$={\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_1$
$={\tan }^{-1}(2022)-{\tan }^{-1}1={\tan }^{-1}2022-\frac{\pi }{4}\text{(option 3)}$
$=\left(\frac{\pi }{2}-{\cot }^{-1}(2022)\right)-\frac{\pi }{4}$
$=\frac{\pi }{4}-{\cot }^{-1}(2022)(\text{option 2)}$
So, the correct option is (B) :

${\left(\frac{4x}{5}+\frac{5}{2x^2}\right)}^9$
Now, $T_{r+1}={}^9{C}_r\cdot {\left(\frac{4x}{5}\right)}^{9-r}{\left(\frac{5}{2x^2}\right)}^r$
$={}^9{C}_r\cdot {\left(\frac{4}{5}\right)}^{9-r}{\left(\frac{5}{2}\right)}^r\cdot x^{9-3r}$
Coefficient of $x^{-6}$ i.e. $9-3r=-6\Rightarrow r=5$
So, Coefficient of $x^{-6}={}^9{C}_5{\left(\frac{4}{5}\right)}^4\cdot {\left(\frac{5}{2}\right)}^5=5040$
So, the correct answer is 5040.

In, ${\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell }}\right)}^9$
$T_{r+1}={}^9{C}_r\frac{{\left(x^{5/2}\right)}^{9-r}}{2^{9-r}}{\left(\frac{-4}{x^{\ell }}\right)}^r$
$=(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r{x}^{\frac{45}{2}-\frac{5r}{2}-r}$
$=45-5r-2lr=0$
$r=\frac{45}{5+21}$ ....(1)
Now, according to the question, $(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r=-84$
$=(-1{)}^r{}^9{C}_r{2}^{3r-9}=21\times 4$
Only natural value of $r$ possible if $3r-9=0$
$r=3$ and ${}^9{C}_3=84$
$\therefore 1=5$ from equation (1)
Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5r}{2}-\mathrm{lr}}$ at $1=5$, gives
$r=5$
$\therefore {}^9{c}_5(-1)\frac{4^5}{2^4}=2^{\alpha }\times \beta$
$=-63\times 2^7$
$\Rightarrow \alpha =7,\beta =-63$
$\therefore \text{value of}|\alpha \ell -\beta |=98$
So, the correct answer is 98.

The correct answer is 98.
In, ${\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell }}\right)}^9$
$T_{r+1}={}^9{C}_r\frac{{\left(x^{5/2}\right)}^{9-r}}{2^{9-r}}{\left(\frac{-4}{x^{\ell }}\right)}^r$
$=(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r{x}^{\frac{45}{2}-\frac{5r}{2}-r}$
$=45-5r-2lr=0$
$r=\frac{45}{5+21}$ ....(1)
Now, according to the question, $(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r=-84$
$=(-1{)}^r{}^9{C}_r{2}^{3r-9}=21\times 4$
Only natural value of $r$ possible if $3r-9=0$
$r=3$ and ${}^9{C}_3=84$
$\therefore 1=5$ from equation (1)
Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5r}{2}-\mathrm{lr}}$ at $1=5$, gives
$r=5$
$\therefore {}^9{c}_5(-1)\frac{4^5}{2^4}=2^{\alpha }\times \beta$
$=-63\times 2^7$
$\Rightarrow \alpha =7,\beta =-63$
$\therefore \text{value of}|\alpha \ell -\beta |=98$