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
 A: 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
$\lfloor v \rfloor = n^2+3n$.
A: $$k<\sqrt{n.(n+1).(n+2).(n+3)+1}=n^2+3n+1$$
$$n.(n+1).(n+2).(n+3)=n^4+6n^3+11n^2+6n>(n^2+3n)^2$$
so $\lfloor k \rfloor =n^2+3n $ 
A: You are using the right approach, we need to establish upper and lower bounds on k. The lower bound needs to be an integer and the upper bound must be no greater than 1 plus the lower bound. This will result in the lower bound being the floor of k. jim established the correct upper bound by adding 1 under the radical.
To see the lower bound, note that $n(n+1)(n+2)(n+3) = n^4 + 6n^3 + 11n^2 + 6n$. Looking at the upper bound, we should try to get a lower bound of $n^2+3n$. So we need $(n^2+3n)^2$ to be a lower bound of $n(n+1)(n+2)(n+3)$. We can see that $(n^2+3n)^2 = n^4 + 6n^3 + 9n^2 < n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$, which establishes our lower bound.
