$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$$
Now by inductive hypothesis:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+k(k+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+k^2(k+1)}=\frac{3k+5}{3k+1}$$
But verification for $n=k+1$ is creating problem.because $k$ is present in denominator too which is hindering use of equation obtained from inductive hypothesis. Please suggest something.

This is also a proof by induction and the result is more general.
The identity
$$\sum_{m=0}^{n}\binom{m}{k}=\binom{n+1}{k+1}  \tag{1}$$
 can be easily proofed by induction using the well known identity of the Pascal Triangle
$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1} \tag{2}$$
$(1)$ says that the sum of this $k$-th degree polynomials in $m$ is a $(k+1)$-st degree polynomial in $n$.
From this it is also simple to proof by induction that for an arbitrary polynomial $p(m)$ of degree $k \in \mathbb{N}$ there is a polynomial $q(m)$ of degree $k+1$ such that 
$$\sum_{m=0}^{n}p(m)=q(m) \tag{3}$$
So 
$$(1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2)({3n+1}) \tag{4})$$
and
$$(1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1))(3n+5) \tag{5}$$
are both polynomials of degree 5. So to see that the polynomials $(4)$ and $(5)$ are equal it is sufficient to show that $(4)$ and $(5)$ have the same values for $n=1,2,3,4,5,6$. These values are $16$, $154$, $700$, $2210$, $5600$ and $12236$.
