Question

If ${\sum }_{k=1}^{n}k(k+1)(k-1)=p{n}^4+q{n}^{3}t{n}^2+sn,$ 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. $\frac{1}{4}$
• E. $\frac{3}{4}$

Answer 1

The correct option is B $-\frac{1}{2}$

Given, $\sum _{k=1}^{n}k(k+1)(k-1)=p{n}^4+q{n}^{3}t{n}^2+sn,$

Therefore, $\sum _{k=1}^n(k^3-k)=p{n}^4+q{n}^3+t{n}^2+sn$

$\Rightarrow {\left(\frac{n(n+1)}{2}\right)}^2-\frac{n(n+1)}{2}=p{n}^4+q{n}^3+t{n}^2+sn$

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

$0-\frac{1}{2}=s$

$\Rightarrow s=-\frac{1}{2}$