Question

# If n > 1 is an integer and $x \ne 0$, then ${\left( {1 + x} \right)^n} - nx - 1$ is divisible by$A.{\text{ }}n{x^3} \\ B.{\text{ }}{n^3}x \\ C.{\text{ }}x \\ D.{\text{ }}nx \\$

Hint: In this question use the concept of binomial theorem such as expansion of ${\left( {1 + x} \right)^n}$ according to binomial expansion to reach the solution of the question.

Given equation is
${\left( {1 + x} \right)^n} - nx - 1$ [where n > 1 and $x \ne 0$]
Then we have to find out the above equation is divisible by.
So as we know the expansion of ${\left( {1 + x} \right)^n}$ according to binomial theorem is
$\Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + {}^n{C_4}{x^4} + .......................... + {}^n{C_n}{x^n}$
Now subtract by $\left( {nx + 1} \right)$ in both sides we have,
$\Rightarrow {\left( {1 + x} \right)^n} - \left( {nx + 1} \right) = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + {}^n{C_4}{x^4} + ........................... + {}^n{C_n}{x^n} - \left( {nx + 1} \right)$
Now as we know that the value of ${}^n{C_0} = 1{text{ \& }}{}^n{C_1} = n$. $\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]$
So on simplifying the above equation we get,
$\Rightarrow {\left( {1 + x} \right)^n} - nx - 1 = 1 + nx + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + {}^n{C_4}{x^4} + .......................... + {}^n{C_n}{x^n} - 1 - nx$
$\Rightarrow {\left( {1 + x} \right)^n} - nx - 1 = {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + {}^n{C_4}{x^4} + ......................... + {}^n{C_n}{x^n}$
Now take ${x^2}$ common from R.H.S we have,
$\Rightarrow {\left( {1 + x} \right)^n} - nx - 1 = {x^2}\left( {{}^n{C_2} + {}^n{C_3}x + {}^n{C_4}{x^2} + ......................... + {}^n{C_n}{x^{n - 2}}} \right)$
So from the above equation it is clear that the given equation is divisible by ${x^2}$ so if the equation is divisible by ${x^2}$ then it is also divisible by x.
So, this is the required answer.
Hence option (c) is correct.

Note: In such types of questions first expand ${\left( {1 + x} \right)^n}$ according to binomial expansion then subtract by $\left( {nx + 1} \right)$ on both sides and simplify then according to property of combination which is stated above the value of ${}^n{C_0} = 1{\text{ & }}{}^n{C_1} = n$ so again simplify and take ${x^2}$ common so whatever is the common, the equation is divisible by common term and if the equation is divisible by ${x^2}$ then it is also divisible by x which is the required answer.