\( (\mathrm{g})^{*} \quad \) Prove that \( (1-x)\left(1+x+x^{2}+\ldots+x^{99}\right)=1-x^{100} \)

To prove that \[ (1 - x)\left(1 + x + x^{2} + \ldots + x^{99}\right) = 1 - x^{100}, \] we can use the properties of geometric series and polynomial multiplication. Here's a step-by-step proof: ### Step 1: Recognize the Sum as a Geometric Series The expression inside the parentheses is a finite geometric series: \[ S = 1 + x + x^{2} + \ldots + x^{99}. \] A geometric series sums terms where each term is a constant multiple (common ratio) of the previous term. The sum of the first \( n \) terms of a geometric series with first term \( a \) and common ratio \( r \) (where \( r \neq 1 \)) is given by: \[ S_n = a \frac{1 - r^{n}}{1 - r}. \] ### Step 2: Apply the Geometric Series Formula For our series: - First term \( a = 1 \) - Common ratio \( r = x \) - Number of terms \( n = 100 \) Applying the formula: \[ S = \frac{1 - x^{100}}{1 - x}. \] This simplifies to: \[ S(1 - x) = 1 - x^{100}. \] ### Step 3: Multiply Out Directly (Alternative Proof) Alternatively, we can directly multiply \( (1 - x) \) by the sum \( S \): \[ (1 - x)(1 + x + x^{2} + \ldots + x^{99}). \] Let's distribute \( (1 - x) \): \[ = 1 \cdot (1 + x + x^{2} + \ldots + x^{99}) - x \cdot (1 + x + x^{2} + \ldots + x^{99}). \] Expanding both parts: \[ = (1 + x + x^{2} + \ldots + x^{99}) - (x + x^{2} + x^{3} + \ldots + x^{100}). \] Notice that most terms cancel out when you subtract: - The \( x \) in the first group cancels with the \( x \) in the second group. - The \( x^{2} \) in the first group cancels with the \( x^{2} \) in the second group. - This pattern continues up to \( x^{99} \). After cancellation, we're left with: \[ 1 - x^{100}. \] ### Conclusion Both methods show that: \[ (1 - x)\left(1 + x + x^{2} + \ldots + x^{99}\right) = 1 - x^{100}. \] Thus, the identity is proven.