If `n > 1` is an integer and `x!=0,` then `(1 +x)^n-nx-1` is divisible by

Options

A. `x^(5)`
B. `x^(2)`
C. `x^(3)`
D. `x^(4)`

Answer

To solve the problem, we need to determine if the expression \((1 + x)^n - nx - 1\) is divisible by a certain factor. Let's break down the solution step by step.

### Step 1: Expand \((1 + x)^n\) using the Binomial Theorem
According to the Binomial Theorem, we can expand \((1 + x)^n\) as follows:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = \binom{n}{0} x^0 + \binom{n}{1} x^1 + \binom{n}{2} x^2 + \ldots + \binom{n}{n} x^n \]
This simplifies to:
\[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2} x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \ldots + x^n \]

### Step 2: Subtract \(nx + 1\) from the expansion
Now, we subtract \(nx + 1\) from the expansion:
\[ (1 + x)^n - nx - 1 = \left(1 + nx + \frac{n(n-1)}{2} x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \ldots + x^n\right) - nx - 1 \]
This simplifies to:
\[ (1 + x)^n - nx - 1 = \frac{n(n-1)}{2} x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \ldots + x^n \]

### Step 3: Factor out \(x^2\)
From the expression obtained, we can see that the first term is \(\frac{n(n-1)}{2} x^2\). Thus, we can factor \(x^2\) out:
\[ (1 + x)^n - nx - 1 = x^2 \left(\frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6} x + \ldots + \frac{1}{x^2} x^{n-2}\right) \]

### Step 4: Analyze divisibility
Since we have factored out \(x^2\), it is clear that the entire expression \((1 + x)^n - nx - 1\) is divisible by \(x^2\).

### Conclusion
Thus, we conclude that \((1 + x)^n - nx - 1\) is divisible by \(x^2\).