Question

If $\sum _{k=1}^{n}k(k+1)(k-1)=p{n}^4+q{n}^3+t{n}^2+sn$, where $p,q,t,s$ are constant, then the correct option(s) is/are

• A. $p=\frac{1}{4}$
• B. $q=\frac{1}{2}$
• C. $t=\frac{1}{4}$
• D. $s=-\frac{1}{2}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $p=\frac{1}{4}$
B $q=\frac{1}{2}$
D $s=-\frac{1}{2}$

$\sum _{k=1}^{n}k(k+1)(k-1)=\sum _{k=1}^n{k}^3-k=\sum _{k=1}^n{k}^3-\sum _{k=1}^{n}k={\left[\frac{n(n+1)}{2}\right]}^2-\left[\frac{n(n+1)}{2}\right]=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}-1]=\frac{n^2+n}{2}\left[\frac{n^2+n-2}{2}\right]=\frac{n^4}{4}+\frac{n^3}{2}-\frac{n^2}{4}-\frac{n}{2}=p{n}^4+q{n}^3+t{n}^2+sn$
$p=\frac{1}{4},q=\frac{1}{2},t=-\frac{1}{4},s=-\frac{1}{2}$