Mean of the numbers 1,2,3,…,n with respective weights
`1^(2) + 1, 2^(2) + 2, 3^(2) + 3,…,n^(2)+n` is

Options

A. `(3n+2)/(2)`
B. `(3n+1)/(4)`
C. `(2n+1)/(3)`
D. `(3n(n+1))/(2(2n +1))`

Answer

To find the mean of the numbers \(1, 2, 3, \ldots, n\) with respective weights \(i^2 + i\), we will follow these steps:

### Step 1: Define the variables
Let \(x_i = i\) (the numbers) and the weights \(w_i = i^2 + i\) for \(i = 1, 2, 3, \ldots, n\).

### Step 2: Write the formula for the mean
The mean can be calculated using the formula:
\[ \text{Mean} = \frac{\sum_{i=1}^{n} w_i \cdot x_i}{\sum_{i=1}^{n} w_i} \]

### Step 3: Calculate the numerator \(\sum_{i=1}^{n} w_i \cdot x_i\)
Substituting \(w_i\) and \(x_i\):
\[ \sum_{i=1}^{n} w_i \cdot x_i = \sum_{i=1}^{n} (i^2 + i) \cdot i = \sum_{i=1}^{n} (i^3 + i^2) \]
This can be separated into two sums:
\[ \sum_{i=1}^{n} i^3 + \sum_{i=1}^{n} i^2 \]

Using the formulas for the sums:
- \(\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2\)
- \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\)

Thus, the numerator becomes:
\[ \sum_{i=1}^{n} w_i \cdot x_i = \left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Calculate the denominator \(\sum_{i=1}^{n} w_i\)
\[ \sum_{i=1}^{n} w_i = \sum_{i=1}^{n} (i^2 + i) = \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i \]
Using the formulas:
- \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\)

Thus, the denominator becomes:
\[ \sum_{i=1}^{n} w_i = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]
Combining these terms gives:
\[ \sum_{i=1}^{n} w_i = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3} \]

### Step 5: Combine the results to find the mean
Now substituting the numerator and denominator into the mean formula:
\[ \text{Mean} = \frac{\left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)(n+2)}{3}} \]

### Step 6: Simplify the expression
To simplify, multiply both numerator and denominator by 6 to eliminate the fractions:
\[ \text{Mean} = \frac{3\left(\frac{n(n+1)}{2}\right)^2 + n(n+1)(2n+1)}{2n(n+1)(n+2)} \]
This simplifies to:
\[ \text{Mean} = \frac{3n^2 + 7n + 2}{4} \]

### Final Result
Thus, the mean of the numbers \(1, 2, 3, \ldots, n\) with respective weights \(i^2 + i\) is:
\[ \text{Mean} = \frac{3n + 1}{4} \]

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