Question

The coefficient of ${x^n}$ in the expansion of ${(1 - x)^{ - 2}}$ is.(A). $\dfrac{{{2^n}}}{{2!}}$(B). $n + 1$(C). $n + 2$(D). $2n$

Hint- In order to find the coefficient of ${x^n}$ first we have to write the binomial expansion of given terms.
The binomial expansion of ${(1 + x)^{ - n}}$ is given as
${(1 + x)^{ - n}} = 1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}}{x^3} + ...........\infty$

Given term ${(1 - x)^{ - 2}}$
We know that the binomial expansion of ${(1 - x)^{ - 2}}$ can be expressed as
$\because {(1 + x)^{ - n}} = 1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}}{x^3} + ...........\infty \\ \Rightarrow {(1 - x)^{ - 2}} = 1 - \left( 2 \right)\left( { - x} \right) + \dfrac{{2\left( {2 + 1} \right)}}{{2!}}{\left( { - x} \right)^2} - \dfrac{{2\left( {2 + 1} \right)\left( {2 + 2} \right)}}{{3!}}{\left( { - x} \right)^3} + ...........\infty \\ \Rightarrow {(1 - x)^{ - 2}} = 1 - 2\left( { - x} \right) + \dfrac{{2 \times 3}}{{2!}}{x^2} - \dfrac{{2 \times 3 \times 4}}{{3!}}\left( { - {x^3}} \right) + ...........\infty \\ \Rightarrow {(1 - x)^{ - 2}} = 1 + 2x + 3{x^2} + 4{x^3} + ...........\infty \\$
Here by observation, we have seen that ${x^0}$ has coefficient 1, coefficient of ${x^1}$ is 2, coefficient of ${x^2}$ is 3. It means the coefficient of ${x^n}$ will be $(n + 1).$
Hence, the coefficient of ${x^n}$ in the expansion of ${(1 - x)^{ - 2}}$ is $(n + 1)$ and the correct answer is option “B”.