Sum of `n` terms of the series `(2n-1)+2(2n-3)+3(2n-5)+...` is

Options

A. `(n(n+1)(2n+1))/6`
B. `(n(n+1)(2n-1))/6`
C. `(n(n-1)(2n-1))/6`
D. none of these

Answer

To find the sum of the series \( (2n-1) + 2(2n-3) + 3(2n-5) + \ldots \), we can follow these steps:

### Step 1: Identify the General Term
The series can be expressed in terms of a general term. The \( r \)-th term of the series can be written as:
\[ T_r = r(2n - (2r - 1)) = r(2n - 2r + 1) \]
This simplifies to:
\[ T_r = r(2n - 2r + 1) = 2nr - 2r^2 + r \]

### Step 2: Write the Sum of the First \( n \) Terms
The sum of the first \( n \) terms, \( S_n \), can be expressed as:
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (2nr - 2r^2 + r) \]

### Step 3: Break Down the Summation
We can separate the summation into three parts:
\[ S_n = \sum_{r=1}^{n} 2nr - \sum_{r=1}^{n} 2r^2 + \sum_{r=1}^{n} r \]

### Step 4: Calculate Each Summation
1. The first summation:
\[ \sum_{r=1}^{n} 2nr = 2n \sum_{r=1}^{n} r = 2n \cdot \frac{n(n+1)}{2} = n(n+1) \]

2. The second summation:
\[ \sum_{r=1}^{n} 2r^2 = 2 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3} \]

3. The third summation:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]

### Step 5: Combine the Results
Now, substituting these results back into the expression for \( S_n \):
\[ S_n = n(n+1) - \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \]

### Step 6: Simplify the Expression
Factoring out \( n(n+1) \):
\[ S_n = n(n+1) \left( 1 - \frac{(2n+1)}{3} + \frac{1}{2} \right) \]

Now, simplifying the expression inside the parentheses:
1. Convert to a common denominator (6):
\[ 1 = \frac{6}{6}, \quad \frac{1}{2} = \frac{3}{6}, \quad \frac{2n+1}{3} = \frac{2(2n+1)}{6} = \frac{4n + 2}{6} \]
So,
\[ 1 - \frac{(2n+1)}{3} + \frac{1}{2} = \frac{6 - (4n + 2) + 3}{6} = \frac{7 - 4n}{6} \]

Thus, we have:
\[ S_n = n(n+1) \cdot \frac{7 - 4n}{6} \]

### Final Result
The sum of the first \( n \) terms of the series is:
\[ S_n = \frac{n(n+1)(7 - 4n)}{6} \]