<s>[INST] $1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.

$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.

Sol:
$P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
$P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.
$P(1): 3$.
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$.
$P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.
We have to prove
$P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.
Taking LHS:
$$\begin{align*}
1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right].
\end{align*}$$
 [/INST] As I understand it you are trying to show that for $n\geq 1$:
$$
1\cdot 3+2\cdot 4+\cdots +n(n+2)=\frac{1}{6}n(n+1)(2n+7).
$$
You first showed it for $n=1$:
$$
3=1\cdot 3=\frac{1}{6}(1)(2)(9)=3
$$
Now we assume that the formula holds for some $n\geq 1$ and we have the statement for $n+1$
$$
3\cdot 1+\cdots +n(n+1)+(n+1)(n+3).
$$
By induction hypothesis the sum of the first $n$ terms is $\frac{1}{6}n(n+1)(2n+7)$.  So
$$
3\cdot 1+\cdots +n(n+1)+(n+1)(n+2)=\frac{1}{6}n(n+1)(2n+7)+(n+1)(n+3).
$$
We may factor out an $n+1$ to get
$$
\frac{1}{6}n(n+1)(2n+7)+(n+1)(n+3)=(n+1)\left(\frac{1}{6}n(2n+7)+n+3\right).
$$
Then factor out a $\frac{1}{6}$ to get
$$
\frac{1}{6}(n+1)(n(2n+7)+6n+18)=\frac{1}{6}(n+1)(2n^2+7n+6n+18).
$$
Finally,
$$
\frac{1}{6}(n+1)(2n^2+7n+6n+18)=\frac{1}{6}(n+1)(n+2)(2(n+1)+7).
$$
</s>