Prove 1+2+3+...+n = n(n+1)/2 by induction

Question

Using the principle of mathematical induction, prove that:

1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

for all positive integers $n$ .

Solution — Step by Step

Let $P(n)$ be the statement:

1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

We need to prove $P(n)$ is true for all $n \in \mathbb{N}$ .

For $n = 1$ :

LHS: $1$

RHS: $\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1$

Since LHS = RHS = 1, P(1) is true. ✓

Assume $P(k)$ is true for some arbitrary positive integer $k$ :

1 + 2 + 3 + \cdots + k = \frac{k(k+1)}{2} \quad \text{...(Inductive Hypothesis)}

We must now prove $P(k+1)$ is true, i.e.,

1 + 2 + 3 + \cdots + k + (k+1) = \frac{(k+1)(k+2)}{2}

Starting with LHS of $P(k+1)$ :

1 + 2 + \cdots + k + (k+1)

Use the inductive hypothesis (replace $1 + 2 + \cdots + k$ ):

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

Factor out $(k+1)$ :

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

This is exactly the RHS of $P(k+1)$ . ✓

Therefore, $P(k+1)$ is true whenever $P(k)$ is true.

Why This Works

Mathematical induction works like a domino chain:

The base case knocks down the first domino ( $P(1)$ is true)
The inductive step shows that if domino $k$ falls, domino $k+1$ also falls

Together, these two facts guarantee ALL dominoes fall — i.e., $P(n)$ is true for all $n \geq 1$ .

The inductive step doesn’t prove anything specific about $k$ — we assume $P(k)$ is true and use that assumption to derive $P(k+1)$ . This is valid because the base case ensures the chain starts, and the inductive step ensures it continues.

Alternative Method — Gauss’s Pairing Trick

Write the sum twice:

S = 1 + 2 + 3 + \cdots + n

S = n + (n-1) + (n-2) + \cdots + 1

Add column by column: each pair sums to $(n+1)$ , and there are $n$ such pairs:

2S = n(n+1) \implies S = \frac{n(n+1)}{2}

This direct method (attributed to young Gauss) is faster but doesn’t constitute a formal induction proof.

Common Mistake

The most common error in induction proofs is circular reasoning in the inductive step — students accidentally assume what they’re trying to prove. The correct structure: “Assume P(k) [this is a hypothesis, not a fact]. Then PROVE P(k+1) using P(k).” You’re proving “IF P(k) then P(k+1)” — the “if” is crucial. Never write “P(k+1) is true, therefore P(k+1) is true.”

CBSE Class 11 expects the full three-part proof: (1) P(1) verification, (2) State the inductive hypothesis clearly, (3) Show P(k) implies P(k+1). Write all three parts explicitly — skipping the base case or the hypothesis statement loses marks.

Question

Solution — Step by Step

Why This Works

Alternative Method — Gauss’s Pairing Trick

Common Mistake

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