If `sum_(k=1)^(n) (sum_(m=1)^(k) m^(2))=an^(4)+bn^(3)+cn^(2)+dn+e`, then

Options

A. `a=(1)/(12)`
B. `b=(1)/(6)`
C. `d=(1)/(4)`
D. e=0

Explanation

To solve the given problem, we need to evaluate the double summation and express it in the form \( an^4 + bn^3 + cn^2 + dn + e \).

### Step-by-Step Solution:

1. **Understand the Double Summation**:
We start with the expression:
\[ \sum_{k=1}^{n} \left( \sum_{m=1}^{k} m^2 \right) \]
The inner summation \( \sum_{m=1}^{k} m^2 \) can be computed using the formula for the sum of squares:
\[ \sum_{m=1}^{k} m^2 = \frac{k(k+1)(2k+1)}{6} \]

2. **Substituting the Inner Sum**:
Substitute the inner sum into the outer summation:
\[ \sum_{k=1}^{n} \left( \frac{k(k+1)(2k+1)}{6} \right) \]
This can be simplified to:
\[ \frac{1}{6} \sum_{k=1}^{n} k(k+1)(2k+1) \]

3. **Expanding the Expression**:
The expression \( k(k+1)(2k+1) \) can be expanded:
\[ k(k+1)(2k+1) = 2k^3 + 3k^2 + k \]
Thus, we can rewrite the summation as:
\[ \frac{1}{6} \left( \sum_{k=1}^{n} (2k^3 + 3k^2 + k) \right) \]

4. **Calculating Each Summation**:
We need to compute each of these summations:
- For \( \sum_{k=1}^{n} k^3 \):
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
- For \( \sum_{k=1}^{n} k^2 \):
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
- For \( \sum_{k=1}^{n} k \):
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

5. **Substituting Back**:
Substitute these results back into the expression:
\[ \frac{1}{6} \left( 2 \left( \frac{n(n+1)}{2} \right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) \]

6. **Simplifying the Expression**:
After substituting and simplifying, we collect like terms:
\[ = \frac{1}{6} \left( \frac{n^4 + 2n^3 + n^2}{2} + \frac{n^3 + 3n^2 + n}{2} + \frac{n(n+1)}{2} \right) \]
This leads to:
\[ = \frac{1}{12} (n^4 + 4n^3 + 7n^2 + 2n) \]

7. **Comparing Coefficients**:
By comparing with \( an^4 + bn^3 + cn^2 + dn + e \):
- \( a = \frac{1}{12} \)
- \( b = \frac{4}{12} = \frac{1}{3} \)
- \( c = \frac{7}{12} \)
- \( d = \frac{2}{12} = \frac{1}{6} \)
- \( e = 0 \)

### Final Result:
Thus, we find:
- \( a = \frac{1}{12} \)
- \( b = \frac{1}{3} \)
- \( c = \frac{7}{12} \)
- \( d = \frac{1}{6} \)
- \( e = 0 \)