(1 + x)^n - nx - 1 is divisible by (where n i | vedclass

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(B) Using the binomial expansion: $(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Subtracting $nx + 1$ from both

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$x^2$

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(B) Using the binomial expansion: $(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Subtracting $nx + 1$ from both sides: $(1 + x)^n - nx - 1 = \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$ Factoring out $x^2$: $(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2!} + \frac{n(n - 1)(n - 2)}{3!}x + \dots \right]$ Since the expression can be written as $x^2$ multiplied by a polynomial in $x$,it is clearly divisible by $x^2$.

$(1 + x)^n - nx - 1$ is divisible by (where $n \in N$):

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