Question

If $\sum _{k=1}^{n}k\left(k+1\right)\left(k-1\right)=p{n}^4+q{n}^3+t{n}^2+sn,$
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, $\sum _{k=1}^{n}k(k+1)(k-1)=p{n}^4+q{n}^{3}t{n}^2+sn$
Therefore, $\sum _{k=1}^n\left(k^3-k\right)=p{n}^4+q{n}^3+t{n}^2+sn$
$\Rightarrow {\left(\frac{n(n+1)}{2}\right)}^2-\frac{n(n+1)}{2}=p{n}^4+q{n}^3+t{n}^2+sn$
On dividing both sides by $n$, we get
$\frac{n(n+1{)}^2}{4}-\frac{(n+1)}{2}=p{n}^3+q{n}^2+tn+s$
Put $n=0$, we get
$0-\frac{1}{2}=s$
$\Rightarrow s=-\frac{1}{2}$