Using the principle of mathematical induction, prove that
`1.2+2.3+3.4+......+n(n+1)=(1)/(3)n(n+1)(n+2)`

Answer

To prove the statement \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1) = \frac{1}{3} n(n+1)(n+2)\) using the principle of mathematical induction, we will follow these steps:

### Step 1: Base Case
We start by checking the base case, \(n = 1\).

**Left Hand Side (LHS)**:
\[ 1 \cdot 2 = 2 \]

**Right Hand Side (RHS)**:
\[ \frac{1}{3} \cdot 1 \cdot (1 + 1) \cdot (1 + 2) = \frac{1}{3} \cdot 1 \cdot 2 \cdot 3 = \frac{6}{3} = 2 \]

Since LHS = RHS for \(n = 1\), the base case holds true.

### Step 2: Induction Hypothesis
Assume the statement is true for some integer \(k\), i.e.,
\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) = \frac{1}{3} k(k+1)(k+2) \]

### Step 3: Induction Step
We need to prove that the statement holds for \(n = k + 1\):
\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) + (k+1)(k+2) = \frac{1}{3} (k+1)(k+2)(k+3) \]

Using the induction hypothesis, we can rewrite the left-hand side:
\[ \frac{1}{3} k(k+1)(k+2) + (k+1)(k+2) \]

Now, factor out \((k+1)(k+2)\):
\[ = (k+1)(k+2) \left( \frac{1}{3} k + 1 \right) \]

Now simplify:
\[ = (k+1)(k+2) \left( \frac{k + 3}{3} \right) \]
\[ = \frac{(k+1)(k+2)(k+3)}{3} \]

This matches the right-hand side for \(n = k + 1\):
\[ \frac{1}{3} (k+1)(k+2)(k+3) \]

### Conclusion
Since both the base case and the induction step have been proven, by the principle of mathematical induction, the statement is true for all natural numbers \(n\).

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