Solveeit Logo
Login

Question

Question: If n>1,then \({{\left( 1+x \right)}^{n}}-nx-1\) is divisible by [a] \(x\) [b] \({{x}^{2}}\) [...

If n>1,then (1+x)n−nx−1{{\left( 1+x \right)}^{n}}-nx-1 is divisible by
[a] xx
[b] x2{{x}^{2}}
[c] x3{{x}^{3}}
[d] x4{{x}^{4}}

Explanation

Solution

Hint: Expand the term (1+x)n{{\left( 1+x \right)}^{n}} using binomial theorem, which states that (x+y)n=∑r=0nnCrxn−ryr{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}. Hence find the lowest degree term in the expression (1+x)n−nx−1{{\left( 1+x \right)}^{n}}-nx-1. If it is r(say), then the expression is divisible by xk,∀k∈[0,r]{{x}^{k}},\forall k\in \left[ 0,r \right]. Hence find which of the options are correct. Alternatively, use the fact that if p(x) is divisible by x, then p(0) = 0, if p(x) is divisible by x2{{x}^{2}}, then p(0)=0,ddxp(x)∣x=0=0p\left( 0 \right)=0,{{\left. \dfrac{d}{dx}p\left( x \right) \right|}_{x=0}}=0 and so on.

Complete step-by-step answer:
We know that
(x+y)n=∑r=0nnCrxn−ryr{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}
Put x= 1 and y = x, we get
(1+x)n=nC0+nC1x+nC2x2+⋯+nCnxn{{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}
We know that nC0=1,nC1=n,nC2=n(n−1)2^{n}{{C}_{0}}=1{{,}^{n}}{{C}_{1}}=n{{,}^{n}}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}
Hence, we have
(1+x)n=1+nx+n(n−1)2x2+⋯+nCnxn{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2}{{x}^{2}}+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}
Subtracting 1+nx from both sides, we get
(1+x)n−nx−1=n(n−1)2x2+nC3x3+⋯+nCnxn{{\left( 1+x \right)}^{n}}-nx-1=\dfrac{n\left( n-1 \right)}{2}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}
Hence, the lowest degree term in the expression (1+x)n−nx−1{{\left( 1+x \right)}^{n}}-nx-1 is n(n−1)2x2\dfrac{n\left( n-1 \right)}{2}{{x}^{2}}
Hence, we have
(1+x)n−1−nx{{\left( 1+x \right)}^{n}}-1-nx is divisible by 1,x and x2{{x}^{2}}
Hence options [a] and [b] are correct.

Note: Alternative Solution:
We know that if p(x)=(x−a)rg(x)p\left( x \right)={{\left( x-a \right)}^{r}}g\left( x \right), then p(a),ddxp(x)∣x=a,d2dx2p(x)∣x=a,⋯ ,dr−1dxr−1p(x)∣x=ap\left( a \right),{{\left. \dfrac{d}{dx}p\left( x \right) \right|}_{x=a}},{{\left. \dfrac{{{d}^{2}}}{d{{x}^{2}}}p\left( x \right) \right|}_{x=a}},\cdots ,{{\left. \dfrac{{{d}^{r-1}}}{d{{x}^{r-1}}}p\left( x \right) \right|}_{x=a}} all are equal to 0.
Now consider p(x)=(1+x)n−nx−1p\left( x \right)={{\left( 1+x \right)}^{n}}-nx-1
We have p(0)=(1)n−0−1=0p\left( 0 \right)={{\left( 1 \right)}^{n}}-0-1=0
Now, p′(x)=n(1+x)n−1−np'\left( x \right)=n{{\left( 1+x \right)}^{n-1}}-n
Hence, we have p′(0)=n(1+0)n−1−n=0p'\left( 0 \right)=n{{\left( 1+0 \right)}^{n-1}}-n=0
Now, we have p′′(x)=n(n−1)(1+x)n−2p''\left( x \right)=n\left( n-1 \right){{\left( 1+x \right)}^{n-2}}
Hence, we have
p′′(0)=n(n−1)p''\left( 0 \right)=n\left( n-1 \right)
Hence, we have p(x)=x2g(x)p\left( x \right)={{x}^{2}}g\left( x \right)
Hence, we have p(x) is divisible by x,x2x,{{x}^{2}}
Hence option [b] is correct.