If `a_1^s in R` then the expression `Sigma_(i = 1)^n (x - a_1)^2` assumes its least. Value at x =

Options

A. `Sigma a_1`
B. `2Sigma a_1`
C. `nSigma a_1`
D. `1/n Sigma a_1`

Answer

To solve the problem, we need to find the value of \( x \) at which the expression

\[ \Sigma_{i=1}^{n} (x - a_i)^2 \]

assumes its least value. This expression represents the sum of squared differences between \( x \) and each \( a_i \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression \( \Sigma_{i=1}^{n} (x - a_i)^2 \) is a quadratic function in terms of \( x \). The function can be expanded as follows:
\[ f(x) = (x - a_1)^2 + (x - a_2)^2 + ... + (x - a_n)^2 \]

2. **Expanding the Expression**:
Expanding each term gives:
\[ f(x) = \Sigma_{i=1}^{n} (x^2 - 2xa_i + a_i^2) = nx^2 - 2x\Sigma_{i=1}^{n} a_i + \Sigma_{i=1}^{n} a_i^2 \]
Here, \( n \) is the number of terms.

3. **Finding the Derivative**:
To find the minimum value, we take the derivative of \( f(x) \):
\[ f'(x) = 2nx - 2\Sigma_{i=1}^{n} a_i \]

4. **Setting the Derivative to Zero**:
To find the critical points, we set the derivative equal to zero:
\[ 2nx - 2\Sigma_{i=1}^{n} a_i = 0 \]
Simplifying this gives:
\[ nx = \Sigma_{i=1}^{n} a_i \]
Thus, we find:
\[ x = \frac{\Sigma_{i=1}^{n} a_i}{n} \]

5. **Conclusion**:
The value of \( x \) that minimizes the expression \( \Sigma_{i=1}^{n} (x - a_i)^2 \) is the average of the \( a_i \) values:
\[ x = \frac{a_1 + a_2 + ... + a_n}{n} \]

### Final Answer:
The expression \( \Sigma_{i=1}^{n} (x - a_i)^2 \) assumes its least value at:
\[ x = \frac{\Sigma_{i=1}^{n} a_i}{n} \]