$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ 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.
 A: As mentioned in the comments, this is probably easier to prove by using formulas for sums of powers. Nevertheless, if you want to proceed by induction, then you could do something like this:
Your formula is equivalent to the statement that $P(n)$ is true for all $n\in\mathbb N$, where
$$
P(n):(3n+1)\sum_{k=1}^n k(k+1)^2-(3n+5)\sum_{k=1}^n k^2(k+1)=0
$$ 
I leave it to you to show that $P(1)$ is true (that does not seem to be your problem). Assume that $P(n)$ is true for some $n\in\mathbb N$. The left-hand side of $P(n+1)$ is then given by
$$
(3n+4)\sum_{k=1}^{n+1}k(k+1)^2-(3n+8)\sum_{k=1}^{n+1}k^2(k+1)
$$
which equals (here we split the sums into the first $n$ terms and the last)
$$
(3n+4)\Bigl[\sum_{k=1}^n k(k+1)^2+(n+1)(n+2)^2\Bigr]-(3n+8)\Bigl[\sum_{k=1}^n k^2(k+1)+(n+1)^2(n+2)\Bigr].
$$
Next, since $P(n)$ was assumed to be true, we cancel $3n+1$ times the sum in the first bracket with $3n+5$ times the sum in the second bracket. Thus, the expression above equals
$$
3\sum_{k=1}^n\bigl[k(k+1)^2-k^2(k+1)\bigr]+(n+1)(n+2)\bigl[(3n+4)(n+2)-(3n+8)(n+1)\bigr]
$$
which simplifies to
$$
3\sum_{k=1}^n k(k+1)-n(n+1)(n+2).
$$
Either we know that this happens to be zero, or we don't. If we don't, we prove it by induction. Thus, let $Q(n)$ be the statement that the expression above is zero. It is easy to verify that $Q(1)$ is true. Next, if $Q(n)$ for some $n\in\mathbb N$ is true, i.e if
$$
3\sum_{k=1}^n k(k+1)=n(n+1)(n+2),
$$
then we find that
$$
3\sum_{k=1}^{n+1}k(k+1)=n(n+1)(n+2)+3(n+1)(n+2)=(n+1)(n+2)(n+3),
$$
which means that $Q(n+1)$ is true. We conclude by induction that $Q(n)$ is true for all $n\in\mathbb N$ and consequently (by induction again) that $P(n)$ is true for all $n\in\mathbb N$.
A: $$1\cdot 2^2 + 2\cdot 3^2 +\cdots+ n\cdot (n+1)^2 =2^3 -2^2 +3^3 - 3^2 +\cdots+(n+1)^3 -(n+1)^2 =\left(\frac{(n+2)(n+1)}{2}\right)^2 -\frac{(n+1)(n+2)(2n+3)}{6} =(n+1)(n+2)\left[\frac{n^2 +3n +2}{4} -\frac{2n+3}{6}\right]= (n+1)(n+2)\left[\frac{3n^2 +5n }{12}\right]=\frac{1}{12} n (n+1) (n+2) (3n+5)$$
$$1^2\cdot 2 + 2^2\cdot 3 +\cdots+ n^2\cdot (n+1) =1^3 +2^3 +3^3  +\cdots+n^3 +1^2 +2^2 +\cdots+n^2 =\left(\frac{n(n+1)}{2}\right)^2 +\frac{n(n+1)(2n+1)}{6} =n(n+1)\left[\frac{n^2 +n}{4} +\frac{2n+1}{6}\right]= n(n+1)\left[\frac{3n^2 +7n +2 }{12}\right]=\frac{1}{12} n (n+1) (n+2) (3n+1)$$
A: 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 $12236$.
