Question

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

Answer 1

Hint: Expand the term ${{\left( 1+x \right)}^{n}}$ using binomial theorem, which states that ${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$. Hence find the lowest degree term in the expression ${{\left( 1+x \right)}^{n}}-nx-1$. If it is r(say), then the expression is divisible by ${{x}^{k}},\forall k\in \left[ 0,r \right]$. 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 ${{x}^{2}}$, then $p\left( 0 \right)=0,{{\left. \dfrac{d}{dx}p\left( x \right) \right|}_{x=0}}=0$ and so on.

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

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