Question

If \(\sum_{k=1}^{n} k(k+1)(k-1) = p n^4 + q n^3 + t n^2 + s n\), where \(p, q, t, s\) are constants, then the value of \(s\) is equal to

• A. \(-\dfrac14\)
• B. \(-\dfrac12\)
• C. \(\dfrac12\)
• D. \( s = -\frac{1}{4} \)

Hint:

Rewrite products into powers before summation.

Answer 1

The Correct Option is A

\(k(k+1)(k-1) = k^3 - k\)
\[ \sum_{k=1}^{n} (k^3 - k) = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} k \]
\[ = \left( \frac{n(n+1)}{2} \right)^2 - \frac{n(n+1)}{2} \]
Expanding and comparing coefficients gives:
\[ s = -\frac{1}{4} \]