Question

If Coefficient of $\sf x^{7}$ in $\sf \bigg(x^2-\dfrac{1}{x} \bigg)^{11}$ , Then Find the value of r

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{an \: expansion} \\ &\sf{\bigg(x^2-\dfrac{1}{x} \bigg)^{11}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{coefficient \: of \: {x}^{7} }  \end{cases}\end{gathered}\end{gathered}$

Concept used : -

$\tt \: In \: the \: expansion \: of \: {(x + a)}^{n} , \: the \: general \: term \: is$

$\tt \: T_{(r + 1)} = \:^n C_r \: {(x)}^{ n-r } {a}^{r}$

$\large\underline{\bold{Solution :- }}$

Now,

• Given series is

$\rm :\longmapsto\:\bigg(x^2-\dfrac{1}{x} \bigg)^{11}$

• The general term is given by

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {( {x}^{2}) }^{11 - r} { \bigg(\dfrac{ - 1}{x} \bigg)}^{r}$

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {x}^{22 - 2r} \dfrac{ {( - 1)}^{r} }{ {x}^{r} }$

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {( - 1)}^{r} {(x)}^{22 - 3r} - - (1)$

Now,

$\rm :\longmapsto\:To \: find \: the \: coefficient \: of \: {x}^{7}$

$\rm :\longmapsto\:put \: 22 - 3r = 7$

$\rm :\longmapsto\:3r = 15$

$\rm :\longmapsto\:r \: = \: 5$

• Now, Substitute r = 5, in equation (1) is

$\rm :\longmapsto\: T_{(5 + 1)} = \:^{11} C_5 \: {( - 1)}^{5} {(x)}^{22 - 3 \times 5}$

$\rm :\longmapsto\: T_{6} = \:^{11} C_5 \: {( - 1)}^{5} {(x)}^{7}$

$\bf\implies \:coefficient \: of \: {x}^{7} = {( - 1)}^{5} \:^{11} C_5$

$\rm :\longmapsto\:\:coefficient \: of \: {x}^{7} = - \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}$

$\rm :\longmapsto\:\:coefficient \: of \: {x}^{7} = \: - \: 462$

Answer 2

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{an \: expansion} \\ &\sf{\bigg(x^2-\dfrac{1}{x} \bigg)^{11}} \end{cases}\end{gathered}\end{gathered}\end{gathered}$

Given−

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{coefficient \: of \: {x}^{7} } \end{cases}\end{gathered}\end{gathered}$

Concept used : -

$\tt \: In \: the \: expansion \: of \: {(x + a)}^{n} , \: the \: general \: term \: isIntheexpansionof(x+a)$

$\large\underline{\bold{Solution :- }}$

Now,

Given series is

$\rm :\longmapsto\:\bigg(x^2-\dfrac{1}{x} \bigg)^{11}:⟼$

The general term is given by

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {( {x}^{2}) }^{11 - r} { \bigg(\dfrac{ - 1}{x} \bigg)}^{r}:⟼T </p><p>(r+1)$

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {x}^{22 - 2r} \dfrac{ {( - 1)}^{r} }{ {x}^{r} }:⟼T$

$\rm :\longmapsto\: T_{(r + 1)} = \:^{11} C_r \: {( - 1)}^{r} {(x)}^{22 - 3r} - - (1):⟼T$

Now,

$\rm :\longmapsto\:To \: find \: the \: coefficient \: of \: {x}^{7}:⟼To find th ecoefficient of x$

$\rm :\longmapsto\:put \: 22 - 3r = 7:⟼put22−3r=7$

$\rm :\longmapsto\:3r = 15:⟼3r=15$

$\rm :\longmapsto\:r \: = \: 5:⟼r=5$

Now, Substitute r = 5, in equation (1) is

$\rm :\longmapsto\: T_{(5 + 1)} = \:^{11} C_5 \: {( - 1)}^{5} {(x)}^{22 - 3 \times 5}:⟼T$

$\rm :\longmapsto\: T_{6} = \:^{11} C_5 \: {( - 1)}^{5} {(x)}^{7}:⟼T$

$\bf\implies \:coefficient \: of \: {x}^{7} = {( - 1)}^{5} \:^{11} C_5⟹coefficientofx$

$\rm :\longmapsto\:\:coefficient \: of \: {x}^{7} = - \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}:⟼coefficient of n x$

$\rm :\longmapsto\:\:coefficient \: of \: {x}^{7} = \: - \: 462:⟼coefficient of x$