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Question: If n > 1 is an integer and \(x \ne 0\), then \({\left( {1 + x} \right)^n} - nx - 1\) is divisible by...

If n > 1 is an integer and x≠0x \ne 0, then (1+x)n−nx−1{\left( {1 + x} \right)^n} - nx - 1 is divisible by
A. nx3 B. n3x C. x D. nx  A.{\text{ }}n{x^3} \\\ B.{\text{ }}{n^3}x \\\ C.{\text{ }}x \\\ D.{\text{ }}nx \\\

Explanation

Solution

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

Complete step-by-step answer:
Given equation is
(1+x)n−nx−1{\left( {1 + x} \right)^n} - nx - 1 [where n > 1 and x≠0x \ne 0]
Then we have to find out the above equation is divisible by.
So as we know the expansion of (1+x)n{\left( {1 + x} \right)^n} according to binomial theorem is
⇒(1+x)n=nC0+nC1x+nC2x2+nC3x3+nC4x4+..........................+nCnxn\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 (nx+1)\left( {nx + 1} \right) in both sides we have,
⇒(1+x)n−(nx+1)=nC0+nC1x+nC2x2+nC3x3+nC4x4+...........................+nCnxn−(nx+1)\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 nC0=1text&nC1=n{}^n{C_0} = 1{text{ \& }}{}^n{C_1} = n. [∵nCr=n!r!(n−r)!]\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]
So on simplifying the above equation we get,
⇒(1+x)n−nx−1=1+nx+nC2x2+nC3x3+nC4x4+..........................+nCnxn−1−nx\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
⇒(1+x)n−nx−1=nC2x2+nC3x3+nC4x4+.........................+nCnxn\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 x2{x^2} common from R.H.S we have,
⇒(1+x)n−nx−1=x2(nC2+nC3x+nC4x2+.........................+nCnxn−2)\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 x2{x^2} so if the equation is divisible by x2{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 (1+x)n{\left( {1 + x} \right)^n} according to binomial expansion then subtract by (nx+1)\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 x2{x^2} common so whatever is the common, the equation is divisible by common term and if the equation is divisible by x2{x^2} then it is also divisible by x which is the required answer.