The mean of n observation is `barX`. If the first observation is increased by `1^(2)`, second by `2^(2)` and so on, then the new mean is

Options

A. `barX+n((n+1)(2n+1))/(6)`
B. `barX+((n+1)(2n+1))/(6)`
C. `barX+(n+1)/(2)`
D. `barX+((n+1))/((4)`

Answer

To find the new mean after modifying the observations, we can follow these steps:

### Step 1: Define the original mean
Let the original observations be \( x_1, x_2, \ldots, x_n \). The mean of these observations is given by:
\[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} x_i \]

### Step 2: Modify the observations
According to the problem, the first observation is increased by \( 1^2 \), the second by \( 2^2 \), and so on, up to the \( n \)-th observation which is increased by \( n^2 \). Therefore, the new observations \( y_i \) can be expressed as:
\[ y_i = x_i + i^2 \quad \text{for } i = 1, 2, \ldots, n \]

### Step 3: Calculate the new mean
The new mean \( \bar{Y} \) is given by:
\[ \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} y_i \]
Substituting for \( y_i \):
\[ \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} (x_i + i^2) = \frac{1}{n} \left( \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} i^2 \right) \]

### Step 4: Separate the sums
We can separate the sums:
\[ \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} x_i + \frac{1}{n} \sum_{i=1}^{n} i^2 \]
This simplifies to:
\[ \bar{Y} = \bar{X} + \frac{1}{n} \sum_{i=1}^{n} i^2 \]

### Step 5: Use the formula for the sum of squares
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting this into our equation for \( \bar{Y} \):
\[ \bar{Y} = \bar{X} + \frac{1}{n} \cdot \frac{n(n + 1)(2n + 1)}{6} \]
This simplifies to:
\[ \bar{Y} = \bar{X} + \frac{(n + 1)(2n + 1)}{6} \]

### Final Result
Thus, the new mean after the modifications is:
\[ \bar{Y} = \bar{X} + \frac{(n + 1)(2n + 1)}{6} \]