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 \( \sum_{r=1}^{n} (2r - 1)^6 \) given that \( \sum_{r=1}^{n} r^6 = f(n) \).

### Step-by-Step Solution:

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

2. **Expanding the Summation**:
We can write the summation as:
\[ \sum_{r=1}^{n} (2r - 1)^6 = (1)^6 + (3)^6 + (5)^6 + \ldots + (2n - 1)^6 \]

3. **Adding and Subtracting Terms**:
To simplify the calculation, we can add and subtract the even terms \( (2)^6, (4)^6, (6)^6, \ldots, (2n)^6 \):
\[ \sum_{r=1}^{n} (2r - 1)^6 + \sum_{r=1}^{n} (2r)^6 - \sum_{r=1}^{n} (2r)^6 \]
This gives us:
\[ \sum_{r=1}^{n} (2r - 1)^6 + \sum_{r=1}^{n} (2r)^6 - \sum_{r=1}^{n} (2r)^6 \]

4. **Rearranging the Terms**:
Now we can rearrange the terms:
\[ \sum_{r=1}^{n} (2r - 1)^6 = \sum_{r=1}^{n} (2r)^6 - \sum_{r=1}^{n} (2r)^6 \]

5. **Calculating the Even Terms**:
The summation \( \sum_{r=1}^{n} (2r)^6 \) can be factored out:
\[ \sum_{r=1}^{n} (2r)^6 = 2^6 \sum_{r=1}^{n} r^6 = 64 f(n) \]

6. **Final Expression**:
Thus, we can express the original summation as:
\[ \sum_{r=1}^{n} (2r - 1)^6 = 64 f(n) - \sum_{r=1}^{n} (2r)^6 \]
Since the last term cancels out, we have:
\[ \sum_{r=1}^{n} (2r - 1)^6 = 64 f(n) \]

### Conclusion:
The final result is:
\[ \sum_{r=1}^{n} (2r - 1)^6 = 64 f(n) \]