Mathematical Induction Notes

The statement to prove is:
$1 + 6 + 11 + \dots + (5n - 1) + 1 = \frac{(n+1)(5n+2)}{2}$

Basis Step: Take the least value of $n$ , which is 0.
- If $n = 0$ , then $P(0)$ is: $1 = \frac{(0+1)(5(0)+2)}{2}$ $1 = \frac{1 \cdot 2}{2}$ $1 = 1$
  - Therefore, $P(0)$ is true.
Inductive Step: Prove that if the statement is true for $k$ , then it will be true for $k+1$ . Prove the implication: If $P(k)$ is true, then $P(k+1)$ is true.

Assume that $P(k)$ is true, where $n = k$ .
$P(k): 1 + 6 + 11 + \dots + (5k + 1) = \frac{(k+1)(5k+2)}{2}$

$P(k+1): 1 + 6 + 11 + \dots + [5(k+1) + 1] = \frac{((k+1)+1)[5(k+1)+2]}{2}$
$1 + 6 + 11 + \dots + (5k + 6) = \frac{(k+2)(5k+7)}{2}$
Need to show that this is true.
The left side is: $1 + 6 + 11 + \dots + (5k + 6)$
The term before $5k + 6$ is $5k + 1$ (since $n = k$ before $n = k+1$ ).
Hypothetically:
$1 + 6 + 11 + \dots + (5k+1) + (5k+6) = \frac{(k+1)(5k+2)}{2} + (5k+6)$
The sum $1 + 6 + 11 + \dots + (5k+1)$ equals $\frac{(k+1)(5k+2)}{2}$ (from the inductive hypothesis).
So,
$\frac{(k+1)(5k+2)}{2} + (5k+6) = \frac{(k+2)(5k+7)}{2}$
$\frac{(k+1)(5k+2) + 2(5k+6)}{2} = \frac{5k^2 + 2k + 5k + 2 + 10k + 12}{2} = \frac{5k^2 + 17k + 14}{2}$
Now, we factor $5k^2 + 17k + 14$ .
$5k^2 + 17k + 14 = (k+2)(5k+7)$
Which is factorable to $(k+2)(5k+7)$ .
Hence,
$\frac{5k^2 + 17k + 14}{2} = \frac{(k+2)(5k+7)}{2}$
Therefore, if $P(k)$ is true, then $P(k+1)$ is true. This means that the implication is true.
Thus, $P$ is true for all $n \geq 0$ .

If you start with $n = 1$ , then $P(1)$ is: $1 + 6 = \frac{(1+1)(5(1)+2)}{2}$ $7 = \frac{2 \cdot 7}{2}$ $7 = 7$
- This is a true statement.
However, if $n=1$ , the left-hand side has two terms, not one.

Given $1 + 6 + 11 + \dots + (5n + 1)$ , the next terms are found by adding 5 each time.
The terms are: 1, 6, 11, 16, 21, …
To go backward, the term before $5n + 1$ is $5n + 1 - 5 = 5n - 4$ .
The term before $5n - 4$ is $5n - 9$ .
The term before $5n - 9$ is $5n - 14$ .