Q42 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics
If $\sum_{k=1}^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n$ where $p, q, t$ and $s$ are constants, then the value of $s$ is equal to
- A$-\frac{1}{4}$
- Correct answerB$-\frac{1}{2}$
- C$\frac{1}{2}$
- D$\frac{1}{4}$
Solution
$\begin{aligned}
\text { Let } S & =\sum_{k=1}^n k(k+1)(k-1) \\
& =\sum_{k=1}^n k^3-k
\end{aligned}$
We know that
$\sum_{r=1}^P r=\frac{P(P+1)}{2}$
and
$\sum_{r=1}^P r^3=\left[\frac{P(P+1)}{2}\right]^2$
$\begin{aligned}
S & =\left[\frac{n(n+1)}{2}\right]^2-\left[\frac{n(n+1)}{2}\right] \\
& =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}-1\right] \\
& =\left(\frac{n^2+n}{2}\right)\left[\frac{n^2+n-2}{2}\right] \\
& =\frac{n^4}{4}+\frac{n^3}{2}-\frac{n^2}{4}-\frac{n}{2}...(i)
\end{aligned}$
Now, by comparing $p n^4+q n^3+t n^2+s n$ with Eq. (i),
$S=-\frac{1}{2}$
Hence, the correct answer is option: 'B'.