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

Question

UpStudy AI · Accepted Answer

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.
$$ (1 - x)\left(1 + x + x^{2} + \ldots + x^{99}\right) = 1 - x^{100} $$

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

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