Question

Prove that ${\left(1+x\right)}^n-nx-1$ is diviisble by $x^2,n\in N$.
Using binomial theorem, prove that $\left(x^n-y^n\right)$ is divisible by $\left(x-y\right)wherex\ne yandn\in N.$

Answer 1

$x^n-y^n$ is divisible by $(x,y)$

${\left(1+x\right)}^n-nx-1$

by using binomial expansion

$\begin{array}{c}\Rightarrow \left[1+nx+\frac{n\left(n-1\right)x^2}{2!}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+.........x^2.x^{n-1}\right]-nx-1\\ \Rightarrow {}^n{c}_2{x}^2{+}^n{c}_3{x}^3+{.........}^n{c}_n{x}^2.x^{n-2}]\\ \Rightarrow x^2\left[{}^n{c}_2{+}^n{c}_3+.......{+}^n{c}_n{x}^{n-2}\right]\end{array}$

Hence ${\left(1+x\right)}^n-nx-1$ is divisible by $x^2$ Proved.