The expression (1 + x)^n - nx - 1 is divisibl | vedclass

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(B) Using the Binomial Theorem,we expand $(1 + x)^n$ as follows:$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$N

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by $x^2$

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(B) Using the Binomial Theorem,we expand $(1 + x)^n$ as follows:$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$Now,subtract $nx + 1$ from the expansion:$(1 + x)^n - nx - 1 = (1 + nx + \frac{n(n - 1)}{2}x^2 + \frac{n(n - 1)(n - 2)}{6}x^3 + \dots) - nx - 1$Simplifying the expression:$(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2} + \frac{n(n - 1)(n - 2)}{6}x + \dots \right]$Since the expression is a multiple of $x^2$,it is divisible by $x^2$.

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

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