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

Question

If $\mathrm{n} > 1$ is an integer and $\mathrm{x} \neq 0$, then $(1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1$ is divisible by, $\mathrm{nx}^3$, $\mathrm{n}^3 \mathrm{x}$, $\mathrm{x}$, $\mathrm{nx}$

MARKS App · Accepted Answer

$\begin{array}{r}\text {Hints: } (1+\mathrm{x})^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 \mathrm{x} 3+\ldots \ldots . \ =1+\mathrm{nx}+\mathrm{x}^2\left({ }^n C_2+{ }^n C_3 \mathrm{x}+\ldots \ldots . .\right) \ (1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1=\mathrm{x}^2\left({ }^n C_2+{ }^n C_3 \mathrm{x}+\ldots \ldots .\right) \end{array}$

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