Question

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

• A. $n{x}^3$
• B. $n^{3}x$
• C. $x$
• D. $nx$

Answer 1

Answer: (C)

$x$

${(1+x)}^n={{}^{n}C}_0+{{}^{n}C}_{1}x+{{}^{n}C}_2{x}^2+{{}^{n}C}_3{x}^3+....$
$.=1+nx+x^2({{}^{n}C}_{2}x+{{}^{n}C}_{3}x+....)$
$\Rightarrow {(1+x)}^n-nx-1=x^2({{}^{n}C}_{2}x+{{}^{n}C}_{3}x+....)$
Hence, it is divisible by $x$