\( (\mathrm{g})^{*} \quad \) 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 concept of a **geometric series** and the **distributive property**. Let's break down the proof step by step. ### Step 1: Recognize the Geometric Series The expression inside the parentheses is a finite geometric series: \[ S = 1 + x + x^{2} + \ldots + x^{99}. \] A geometric series has the form: \[ S = a + ar + ar^{2} + \ldots + ar^{n-1}, \] where: - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. In our case: - \( a = 1 \), - \( r = x \), - \( n = 100 \). ### Step 2: Use the Finite Geometric Series Formula The sum of a finite geometric series is given by: \[ S = \frac{a(1 - r^{n})}{1 - r}. \] Plugging in our values: \[ S = \frac{1(1 - x^{100})}{1 - x} = \frac{1 - x^{100}}{1 - x}. \] ### Step 3: Multiply Both Sides by \(1 - x\) Now, multiply both sides of the equation by \(1 - x\): \[ (1 - x)S = 1 - x^{100}. \] Substituting back the expression for \(S\): \[ (1 - x)\left(1 + x + x^{2} + \ldots + x^{99}\right) = 1 - x^{100}. \] ### Alternative Proof: Direct Distribution (Telescoping Method) Alternatively, we can directly multiply \(1 - x\) with the series term by term: \[ \begin{align*} (1 - x)(1 + x + x^{2} + \ldots + x^{99}) &= 1 \cdot (1 + x + x^{2} + \ldots + x^{99}) \\ &\quad - x \cdot (1 + x + x^{2} + \ldots + x^{99}) \\ &= (1 + x + x^{2} + \ldots + x^{99}) \\ &\quad - (x + x^{2} + x^{3} + \ldots + x^{100}). \end{align*} \] When we subtract the two series, almost all terms cancel out (this is known as a telescoping series): \[ \begin{align*} &1 + \underline{x} + \underline{x^{2}} + \ldots + \underline{x^{99}} \\ &- \underline{x} - \underline{x^{2}} - \underline{x^{3}} - \ldots - x^{100} \\ &= 1 - x^{100}. \end{align*} \] The underlined terms cancel each other, leaving only: \[ 1 - x^{100}. \] ### Conclusion Both methods conclusively show that: \[ (1 - x)\left(1 + x + x^{2} + \ldots + x^{99}\right) = 1 - x^{100}. \] This identity is a specific case of the more general formula for the sum of a finite geometric series.