If n 1 is an integer and x neq 0, then (1+x | vedclass

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(C) Using the Binomial Theorem, we have: $(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + {^nC_3}x^3 + \dots + {^nC_n}x^n$ Since ${^nC_0} = 1$ and ${^nC_1

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$x^2$

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(C) Using the Binomial Theorem, we have: $(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + {^nC_3}x^3 + \dots + {^nC_n}x^n$ Since ${^nC_0} = 1$ and ${^nC_1} = n$, we can write: $(1+x)^n = 1 + nx + {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$ Now, subtract $nx$ and $1$ from both sides: $(1+x)^n - nx - 1 = {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$ Factoring out $x^2$ from the right side: $(1+x)^n - nx - 1 = x^2 ({^nC_2} + {^nC_3}x + \dots + x^{n-2})$ Thus, the expression is divisible by $x^2$.

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

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