# Mathematical Induction Proof with Factorials | Principles of Mathematical Induction

Prove $2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (n^2+1)n! = n(n+1)!$.

### Step 1

Show it is true for $n=1$.

\begin{align} &\text{LHS} = (1^2+1) \times 1! = 2 \\ &\text{RHS} = 1 \times (1+1)! = 2 \\ &\text{LHS} = \text{RHS} \\ &\text{Therefore it is true for } n=1 \end{align}

### Step 2

Assume that it is true for $n=k$, that is, $2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (k^2+1)k! = k(k+1)!$

### Step 3

Show it is true for $n=k+1$, that is, $2 \times 1! + \cdots + (k^2+1)k! + (k^2+2k+2)(k+1)! = (k+1)(k+2)!$

\begin{align} \text{LHS} &= 2 \times 1! + \cdots + (k^2+1)k! + (k^2+2k+2)(k+1)! \\ &= k(k+1)!+(k^2+2k+2)(k+1)! \\ &= \left[k+\left(k^2+2k+2\right) \right](k+1)! \\ &= \left(k^2+3k+2\right)(k+1)! \\ &= (k+1)(k+2)(k+1)! \\ &= (k+1)(k+2)! \\ &= \text{RHS} \end{align}

Therefore it is true for $n=k+1$.

Therefore the statement is true for all integers $n \ge 1$.