If `x gt -1` , then the statement `(1+x)^n gt 1 +nx` is true for

Options

A. all `n in N`
B. all n `gt` 2
C. all `n gt 1` provided `x ne 0`
D. None of these

Answer

To prove the statement \( (1+x)^n > 1 + nx \) for \( x > -1 \), we will use the Principle of Mathematical Induction.

### Step 1: Base Case
We start with the base case where \( n = 1 \).

\[ (1+x)^1 = 1 + x \]
\[ 1 + 1 \cdot x = 1 + x \]

Here, both sides are equal, so the statement does not hold for \( n = 1 \).

### Step 2: Base Case for \( n = 2 \)
Next, we check the case for \( n = 2 \).

\[ (1+x)^2 = 1 + 2x + x^2 \]
\[ 1 + 2x \]

We need to check if \( 1 + 2x + x^2 > 1 + 2x \).

Subtracting \( 1 + 2x \) from both sides gives:

\[ x^2 > 0 \]

This is true for all \( x \neq 0 \). Thus, the statement holds for \( n = 2 \) provided \( x \neq 0 \).

### Step 3: Inductive Step
Assume the statement is true for some \( n = k \), i.e.,

\[ (1+x)^k > 1 + kx \]

Now we need to prove it for \( n = k + 1 \):

\[ (1+x)^{k+1} = (1+x)^k(1+x) \]

By the inductive hypothesis, we have:

\[ (1+x)^{k+1} > (1 + kx)(1+x) \]

Expanding the right-hand side:

\[ (1 + kx)(1+x) = 1 + kx + x + kx^2 = 1 + (k+1)x + kx^2 \]

We need to show:

\[ (1+x)^{k+1} > 1 + (k+1)x \]

This simplifies to showing:

\[ (1+x)^{k+1} > 1 + (k+1)x + kx^2 \]

Since \( kx^2 \) is always non-negative for \( x > -1 \), we can conclude that:

\[ (1+x)^{k+1} > 1 + (k+1)x \]

### Conclusion
By the principle of mathematical induction, we have shown that \( (1+x)^n > 1 + nx \) holds true for all \( n \geq 2 \) and \( x > -1 \) (with the condition that \( x \neq 0 \) for \( n = 1 \)).