If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdot (n^2+3n)+1-1 = (n^2+3n+1)^2-1<(n^2+3n)^2$
Now I Did not Understand How can i solve further
Help Required.
Thanks

We want to determine
$\lfloor v \rfloor$,
where
$v = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$.
Let $x = n+3/2$,
so
$x^2 = n^2+3n+9/4$.
Then
$\begin{align}
v^2
&=(x-3/2)(x-1/2)(x+1/2)(x+3/2)\\
&=(x^2-(3/2)^2)(x^2-(1/2)^2)\\
&=x^4-5x^2/2+9/16\\
&=(x^2-5/4)^2-25/16+9/16\\
&=((n^2+3n+9/4)-5/4)^2-1\\
&=(n^2+3n+1)^2-1\\
\end{align}
$
Therefore
$v < n^2+3n+1$
and
$v > n^2+3n$
(since
$u^2-1 > (u-1)^2$
for $u > 1$)
so
$\lfloor v \rfloor = n^2+3n$.
