Question

Let the Sum of $n.n+(n-1).(n+1)+(n-2).(n+2)+.....+1.(2n-1)$ is $\frac{1}{k}n(n+1)(mn-1)$. FInd $k-m$.

Answer 1

$n.n+(n-1).(n+1)+(n-2).(n+2)+.....+1.(2n-1)$,
Expanding the above equation, we get
$n.n+(n-1).(n+1)+(n-2).(n+2)+.....+(n-(n-1\left)\right).(n+(n-1\left)\right)$
$\Rightarrow n.n^2-(1^2+2^2+\cdots +(n-1{)}^2$
$\Rightarrow n^3-\sum _{n=1}^n(n-1{)}^2=n^3-\frac{n(n-1)(2n-1)}{6}$
$\Rightarrow \frac{1}{6}n(n+1)(4n-1)=\frac{1}{k}n(n+1)(mn-1)$
$m=4,k=6$, then $k-m=6-4=2$