Let a, b, c, d be real numbers such that
`sum_(k=1)^(n) (ak^(3) + bk^(2) + ck + d) = n^(4)`
or every natural number n. Then | a | + | b | + | C | + | d | is equal to

Options

A. 15
B. 16
C. 31
D. 32

Answer

To solve the problem, we need to find the values of \( a, b, c, d \) such that:

\[ \sum_{k=1}^{n} (ak^3 + bk^2 + ck + d) = n^4 \]

for every natural number \( n \).

### Step 1: Expand the summation
We can break down the summation into separate parts:

\[ \sum_{k=1}^{n} ak^3 + \sum_{k=1}^{n} bk^2 + \sum_{k=1}^{n} ck + \sum_{k=1}^{n} d \]

Using the known formulas for these summations:
- \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} 1 = n \)

We can substitute these into our equation:

\[ a \left(\frac{n(n+1)}{2}\right)^2 + b \frac{n(n+1)(2n+1)}{6} + c \frac{n(n+1)}{2} + dn = n^4 \]

### Step 2: Simplify the equation
Now, we can simplify each term:

1. The first term becomes:
\[ a \cdot \frac{n^2(n+1)^2}{4} = \frac{a}{4} n^4 + \frac{a}{2} n^3 + \frac{a}{4} n^2 \]

2. The second term becomes:
\[ b \cdot \frac{n(n+1)(2n+1)}{6} = \frac{b}{6} (2n^3 + 3n^2 + n) = \frac{b}{3} n^3 + \frac{b}{2} n^2 + \frac{b}{6} n \]

3. The third term becomes:
\[ c \cdot \frac{n(n+1)}{2} = \frac{c}{2} n^2 + \frac{c}{2} n \]

4. The fourth term becomes:
\[ dn \]

### Step 3: Combine all terms
Combining all these, we have:

\[ \left(\frac{a}{4}\right) n^4 + \left(\frac{a}{2} + \frac{b}{3}\right) n^3 + \left(\frac{a}{4} + \frac{b}{2} + \frac{c}{2}\right) n^2 + \left(\frac{b}{6} + \frac{c}{2} + d\right) n = n^4 \]

### Step 4: Set coefficients equal
For this equation to hold for all \( n \), the coefficients of \( n^4, n^3, n^2, n \) must be equal on both sides:

1. From \( n^4 \):
\[ \frac{a}{4} = 1 \implies a = 4 \]

2. From \( n^3 \):
\[ \frac{a}{2} + \frac{b}{3} = 0 \implies 2 + \frac{b}{3} = 0 \implies b = -6 \]

3. From \( n^2 \):
\[ \frac{a}{4} + \frac{b}{2} + \frac{c}{2} = 0 \implies 1 - 3 + \frac{c}{2} = 0 \implies \frac{c}{2} = 2 \implies c = 4 \]

4. From \( n \):
\[ \frac{b}{6} + \frac{c}{2} + d = 0 \implies -1 + 2 + d = 0 \implies d = -1 \]

### Step 5: Calculate the final answer
Now we have:
- \( a = 4 \)
- \( b = -6 \)
- \( c = 4 \)
- \( d = -1 \)

We need to find \( |a| + |b| + |c| + |d| \):

\[ |a| + |b| + |c| + |d| = |4| + |-6| + |4| + |-1| = 4 + 6 + 4 + 1 = 15 \]

Thus, the final answer is:

\[ \boxed{15} \]