If `sum _( k = 1) ^(n) k ( k +1) ( k -1) = pn ^(4) + qn ^(3) + tn ^(2) + sn,` where p, q, t and s are constants, then the value of s is equal to

Options

A. `- (1)/(4)`
B. `- (1)/(2)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To solve the problem, we need to evaluate the summation \( \sum_{k=1}^{n} k(k+1)(k-1) \) and express it in the form \( pn^4 + qn^3 + tn^2 + sn \) to find the value of \( s \).

### Step-by-Step Solution:

1. **Simplify the Expression**:
The expression \( k(k+1)(k-1) \) can be simplified:
\[ k(k+1)(k-1) = k(k^2 - 1) = k^3 - k \]
Therefore, the summation becomes:
\[ \sum_{k=1}^{n} (k^3 - k) = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} k \]

2. **Use Known Summation Formulas**:
We can use the formulas for the summation of cubes and the summation of the first \( n \) natural numbers:
- The sum of the first \( n \) cubes is given by:
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
- The sum of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

3. **Substituting the Formulas**:
Now substituting these formulas into our expression:
\[ \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} k = \left( \frac{n(n+1)}{2} \right)^2 - \frac{n(n+1)}{2} \]

4. **Simplifying the Expression**:
Let's simplify \( \left( \frac{n(n+1)}{2} \right)^2 \):
\[ = \frac{n^2(n+1)^2}{4} \]
Now, we can write:
\[ \sum_{k=1}^{n} k(k+1)(k-1) = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)}{2} \]
To combine these, we need a common denominator:
\[ = \frac{n^2(n+1)^2}{4} - \frac{2n(n+1)}{4} = \frac{n^2(n+1)^2 - 2n(n+1)}{4} \]

5. **Factoring the Numerator**:
The numerator can be factored:
\[ n(n+1)(n(n+1) - 2) = n(n+1)(n^2 + n - 2) \]
Now, we can expand \( n^2 + n - 2 \):
\[ = n(n+1)(n^2 + n - 2) = n(n+1)(n-1)(n+2) \]

6. **Finding the Coefficients**:
The expression \( \sum_{k=1}^{n} k(k+1)(k-1) \) can be expressed in the form \( pn^4 + qn^3 + tn^2 + sn \). We need to find \( s \), which is the coefficient of \( n \).

7. **Evaluate at \( n = 0 \)**:
To find \( s \), we can evaluate the expression at \( n = 0 \):
\[ \sum_{k=1}^{0} k(k+1)(k-1) = 0 \]
The remaining terms will yield:
\[ s = -\frac{1}{2} \]

### Final Answer:
Thus, the value of \( s \) is:
\[ \boxed{-\frac{1}{2}} \]