Prove the statement \[ 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n+1) \] by using the principle of mathematical induction.

Proof:

Let \[ P(n) \] be the statement:

\[ 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n+1) \]

Basic Step:

For \[ n=1 \]

\[ \text{LHS} = 1 \];
\[ \text{RHS} = \frac{1}{2} \times 1 \times (1+1) \]
\[ = \frac{1}{2} \times 2 \]
\[ = 1 \]

\[ \therefore P(1) \] is true

Inductive Step:

Assume \[ P(k) \] is true.

i.e., \[ 1 + 2 + 3 + \dots + k = \frac{1}{2} k (k+1) \]

T.P \[ P(k+1) \text{ is true.} \]

i.e., T.P \[ 1 + 2 + 3 + \dots + k + (k+1) = \frac{1}{2} (k+1)(k+2) \]

\[ \text{LHS}: \]

\[ 1 + 2 + 3 + \dots + k + (k+1) = \frac{1}{2} k(k+1)+k+1 \]

\[= \frac{k(k+1)+2(k+1)}{2} \]

\[= \frac{1}{2}[k^2 + k + 2k + 2] \]

\[= \frac{1}{2}[k^2 + 3k + 2] \]

\[= \frac{1}{2}[(k + 1)(k + 2)] \]

\[= \text{RHS} \]

\[ \therefore P(k + 1) \] is true

\[ \therefore P(n) \] is true

\[ \therefore 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n+1) \forall n \]