Let `sum_(r=1)^(n) r^(6)=f(n)," then "sum_(n=1)^(n) (2r-1)^(6)` is equal to

Options

A. `f(n)-64f((n+1)/(2))` n is odd
B. `f(n)-64f((n-1)/(2))` n is odd
C. `f(n)-64f((n)/(2))`, n is even
D. none of these

Explanation

To solve the problem, we need to evaluate the summation \( \sum_{r=1}^{n} (2r-1)^6 \) in terms of \( f(n) = \sum_{r=1}^{n} r^6 \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to evaluate \( \sum_{r=1}^{n} (2r-1)^6 \). The term \( (2r-1) \) represents the odd numbers from 1 to \( 2n-1 \).

2. **Expanding the Summation**:
We can rewrite the summation:
\[ \sum_{r=1}^{n} (2r-1)^6 = (1)^6 + (3)^6 + (5)^6 + \ldots + (2n-1)^6 \]
This is the sum of the sixth powers of the first \( n \) odd numbers.

3. **Relating to Even Numbers**:
The sixth powers of the first \( 2n \) integers can be expressed as:
\[ \sum_{r=1}^{2n} r^6 \]
This includes both odd and even integers. The even integers from 1 to \( 2n \) can be expressed as \( 2, 4, 6, \ldots, 2n \).

4. **Separating Odd and Even Terms**:
The sum of the sixth powers of the even integers can be expressed as:
\[ \sum_{r=1}^{n} (2r)^6 = 2^6 \sum_{r=1}^{n} r^6 = 64 f(n) \]
where \( f(n) = \sum_{r=1}^{n} r^6 \).

5. **Combining the Results**:
Now, we can express the sum of the sixth powers of the first \( 2n \) integers as:
\[ \sum_{r=1}^{2n} r^6 = \sum_{r=1}^{n} (2r-1)^6 + \sum_{r=1}^{n} (2r)^6 \]
Substituting the expressions we have:
\[ f(2n) = \sum_{r=1}^{n} (2r-1)^6 + 64 f(n) \]

6. **Isolating the Desired Summation**:
Rearranging gives us:
\[ \sum_{r=1}^{n} (2r-1)^6 = f(2n) - 64 f(n) \]

### Final Result:
Thus, the summation \( \sum_{r=1}^{n} (2r-1)^6 \) is equal to:
\[ f(2n) - 64 f(n) \]