Question

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

• A. $ - \frac{1}{4}$
• B. $ - \frac{1}{2}$
• C. $ \frac{1}{2}$
• D. $ \frac{1}{4}$

Answer 1

Answer: (B)

Given, $\displaystyle\sum_{ k =1}^{ n } k ( k +1)( k -1)= pn ^{4}+ qn ^{3} tn ^{2}+ sn$
Therefore, $\displaystyle\sum_{ k =1}^{ n }\left( k ^{3}- k \right)= pn ^{4}+ qn ^{3}+ tn ^{2}+ sn$
$\Rightarrow \left(\frac{ n ( n +1)}{2}\right)^{2}-\frac{ n ( n +1)}{2}= pn ^{4}+ qn ^{3}+ tn ^{2}+ sn$
On dividing both sides by $n$, we get
$\frac{ n ( n +1)^{2}}{4}-\frac{( n +1)}{2}= pn ^{3}+ qn ^{2}+ tn + s$
Put $n =0$, we get
$0-\frac{1}{2}= s$
$\Rightarrow s =-\frac{1}{2}$