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Question

Answers

$

A.{\text{ }}n{x^3} \\

B.{\text{ }}{n^3}x \\

C.{\text{ }}x \\

D.{\text{ }}nx \\

$

Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.