$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ 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:
$$\begin{align*}
1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right].
\end{align*}$$

You used the wrong formula in your induction hypothesis. You are assuming that
$$1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$$
but in your inductive argument, you wrote
$$1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3) = \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3).$$
That is, you substituted $1\cdot 3 + 2\cdot 4 + \cdots + k(k+2)$ with the formula for $k+1$, not the formula for $k$.
