1.18. By applying formulas (1.14.6) and (1.9.1) show that L (Xi - X)2 i= 1 
can be rewritten in the following three forms which are often 
useful in computation: 
a) LX? - nx2 , 
c) LX? - x(L x;) .

To show that the sum of squares \( \sum_{i=1}^{n} (X_i - \bar{X})^2 \) can be rewritten in the specified forms, we’ll apply formulas (1.14.6) and (1.9.1). For this explanation, let's denote:

- \( X_i \) as the \( i \)-th observation,
- \( \bar{X} \) as the mean of \( X_i \) values, i.e., \( \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \),
- \( n \) as the number of observations.

### Part (a): Show that
\[
\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2
\]

#### Step-by-Step Solution:

1. **Expand the left-hand side**: Start with the definition of the sum of squared deviations:
   \[
   \sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} \left( X_i^2 - 2 X_i \bar{X} + \bar{X}^2 \right).
   \]

2. **Separate the terms**: Distribute the summation:
   \[
   = \sum_{i=1}^{n} X_i^2 - 2 \sum_{i=1}^{n} X_i \bar{X} + \sum_{i=1}^{n} \bar{X}^2.
   \]

3. **Simplify each term**:
   - The first term, \( \sum_{i=1}^{n} X_i^2 \), remains as it is.
   - The second term becomes \( -2 \bar{X} \sum_{i=1}^{n} X_i \). Since \( \sum_{i=1}^{n} X_i = n \bar{X} \), this term simplifies to \( -2 \bar{X} \cdot n \bar{X} = -2 n \bar{X}^2 \).
   - The third term, \( \sum_{i=1}^{n} \bar{X}^2 \), simplifies to \( n \bar{X}^2 \) because \( \bar{X}^2 \) is constant with respect to \( i \).

4. **Combine terms**:
   \[
   \sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - 2 n \bar{X}^2 + n \bar{X}^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2.
   \]

This proves part (a).

---

### Part (c): Show that
\[
\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - \bar{X} \sum_{i=1}^{n} X_i.
\]

1. **Use the result from part (a)**: From part (a), we know that
   \[
   \sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2.
   \]

2. **Substitute \( n \bar{X} \) with \( \sum_{i=1}^{n} X_i \)**: Recall that \( \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \), so \( n \bar{X} = \sum_{i=1}^{n} X_i \).

3. Rewrite \( n \bar{X}^2 \):
   \[
   n \bar{X}^2 = \bar{X} \sum_{i=1}^{n} X_i.
   \]

4. Substitute into the formula:
   \[
   \sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - \bar{X} \sum_{i=1}^{n} X_i.
   \]

This completes the proof for part (c).

---

Let me know if you’d like further details or have any questions about these derivations.

---

**Relative Questions:**
1. What is the geometric interpretation of \( \sum_{i=1}^{n} (X_i - \bar{X})^2 \) in statistics?
2. How can these rewritten forms be useful for computational efficiency?
3. What is the meaning of \( \bar{X} \) in terms of probability if \( X_i \) represents a sample?
4. How does this approach change if \( X_i \) values are weighted?
5. What role does the variance play in relation to these forms?

**Tip:** Remember, expanding and simplifying terms step-by-step helps avoid mistakes when dealing with summation formulas.

By applying formulas (1.14.6) and (1.9.1) show that ∑(Xi - X̄)² can be rewritten in the following three forms which are often useful in computation: a) ∑Xi² - nX̄², c) ∑Xi² - X̄∑Xi.

Explore how to express the sum of squared deviations, ∑(Xi - X̄)², in alternative forms useful for computational efficiency. This solution demonstrates each step, applying summation and mean properties to derive forms such as ∑Xi² - nX̄². Ideal for college-level statistics or mathematics students.

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