Let `sum_(r=1)^(n)r^(4)=f(n)," then " sum_(r=1)^(n) (2r-1)^(4)` is equal to

Options

A. f(2n)-16f(n)
B. f(2n)-7f(n)
C. f(2n-1)-8f(n)
D. none of these

Answer

To solve the problem, we need to find the expression for the summation \( \sum_{r=1}^{n} (2r-1)^4 \) given that \( \sum_{r=1}^{n} r^4 = f(n) \).

### Step-by-step Solution:

1. **Understanding the Summation**:
We need to evaluate \( \sum_{r=1}^{n} (2r-1)^4 \). The term \( (2r-1) \) represents the sequence of odd numbers starting from 1 up to \( 2n-1 \).

2. **Expanding the Expression**:
We can expand \( (2r-1)^4 \) using the binomial theorem:
\[ (2r-1)^4 = \sum_{k=0}^{4} \binom{4}{k} (2r)^{4-k} (-1)^k \]
This gives us:
\[ (2r-1)^4 = 16r^4 - 32r^3 + 24r^2 - 8r + 1 \]

3. **Substituting into the Summation**:
Now we substitute this expansion into our summation:
\[ \sum_{r=1}^{n} (2r-1)^4 = \sum_{r=1}^{n} (16r^4 - 32r^3 + 24r^2 - 8r + 1) \]

4. **Separating the Summation**:
We can separate the summation into individual parts:
\[ = 16 \sum_{r=1}^{n} r^4 - 32 \sum_{r=1}^{n} r^3 + 24 \sum_{r=1}^{n} r^2 - 8 \sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 \]

5. **Using Known Formulas**:
We can use the known formulas for the summations:
- \( \sum_{r=1}^{n} r^4 = f(n) \)
- \( \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \)
- \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
- \( \sum_{r=1}^{n} 1 = n \)

6. **Substituting the Formulas**:
Substitute these formulas into our expression:
\[ = 16f(n) - 32 \left( \frac{n(n+1)}{2} \right)^2 + 24 \left( \frac{n(n+1)(2n+1)}{6} \right) - 8 \left( \frac{n(n+1)}{2} \right) + n \]

7. **Simplifying the Expression**:
Now we simplify each term:
- The first term remains \( 16f(n) \).
- The second term becomes \( -8n^2(n+1)^2 \).
- The third term simplifies to \( 4n(n+1)(2n+1) \).
- The fourth term simplifies to \( -4n(n+1) \).
- The last term is simply \( n \).

8. **Final Expression**:
Combine all the terms to get the final expression for \( \sum_{r=1}^{n} (2r-1)^4 \):
\[ \sum_{r=1}^{n} (2r-1)^4 = 16f(n) - 8n^2(n+1)^2 + 4n(n+1)(2n+1) - 4n(n+1) + n \]

### Final Answer:
\[ \sum_{r=1}^{n} (2r-1)^4 = 16f(n) - 8n^2(n+1)^2 + 4n(n+1)(2n+1) - 4n(n+1) + n \]