   Question Answers

# 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”.

Note- In order to solve these types of questions, we need to remember the formula of binomial expansion and with the help of this any equation of similar kind as above can be expanded and the coefficient can be similarly calculated. Few properties of binomial expansion are the binomial coefficients which are equidistant from the beginning and from the ending are equal.

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