Sum of the series `S_(n) =(n) (n) + (n-1) (n+1) + (n-2) (n+2) + …+ 1(2n-1)` is

Options

A. `(1)/(6) n (n+1) (n+2)`
B. `(1)/(3) n^(3)-n^(2)`
C. `n^(3)`
D. none of these

Answer

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

### Step 1: Write the series in a more manageable form
The series can be expressed as:
\[ S_n = n^2 + (n-1)(n+1) + (n-2)(n+2) + \ldots + 1(2n-1) \]
Notice that each term can be rewritten using the difference of squares:
\[ (n-k)(n+k) = n^2 - k^2 \]
for \( k = 0, 1, 2, \ldots, n-1 \).

### Step 2: Expand the series
Using the above identity, we can rewrite the series:
\[ S_n = n^2 + (n^2 - 1^2) + (n^2 - 2^2) + \ldots + (n^2 - (n-1)^2) \]
This simplifies to:
\[ S_n = n^2 + n^2(n) - (1^2 + 2^2 + \ldots + (n-1)^2) \]

### Step 3: Count the number of terms
There are \( n \) terms in the series, and each term contributes \( n^2 \) to the sum:
\[ S_n = n \cdot n^2 - (1^2 + 2^2 + \ldots + (n-1)^2) \]

### Step 4: Use the formula for the sum of squares
The sum of squares formula is:
\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} \]
For our case, \( m = n-1 \):
\[ 1^2 + 2^2 + \ldots + (n-1)^2 = \frac{(n-1)n(2(n-1)+1)}{6} = \frac{(n-1)n(2n-1)}{6} \]

### Step 5: Substitute back into the equation
Now substituting this back into our expression for \( S_n \):
\[ S_n = n^3 - \frac{(n-1)n(2n-1)}{6} \]

### Step 6: Simplify the expression
Now we can simplify:
\[ S_n = n^3 - \frac{(n^2 - n)(2n - 1)}{6} \]
This can be expressed as:
\[ S_n = n^3 - \frac{2n^3 - n^2 - 2n^2 + n}{6} \]
Combining the terms gives:
\[ S_n = n^3 - \frac{2n^3 - 3n^2 + n}{6} \]
To combine the fractions:
\[ S_n = \frac{6n^3 - (2n^3 - 3n^2 + n)}{6} \]
\[ = \frac{6n^3 - 2n^3 + 3n^2 - n}{6} \]
\[ = \frac{4n^3 + 3n^2 - n}{6} \]

### Final Result
Thus, the sum of the series is:
\[ S_n = \frac{4n^3 + 3n^2 - n}{6} \]