Induction proof for the sum of the first n squares

Expert intro: This proof uses mathematical induction, one of the most important techniques for proving identities about positive integers.

Detailed walkthrough

State the proposition clearly

We want to prove that for every positive integer $n$,

$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$

Mathematical induction has two parts: the base case and the inductive step.

Base case

For $n=1$, the left-hand side is

$1^2=1,$

and the right-hand side is

$\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot2\cdot3}{6}=1.$

So the formula is true when $n=1$.

Inductive step

Assume the statement is true for some positive integer $k$. That is, assume

$1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}.$

We must show it is true for $k+1$.

Start with the left-hand side for $k+1$:

$1^2+2^2+\cdots+k^2+(k+1)^2.$

Use the inductive hypothesis:

$=\frac{k(k+1)(2k+1)}{6}+(k+1)^2.$

Factor out $(k+1)$:

$=(k+1)\left(\frac{k(2k+1)}{6}+k+1\right).$

Put everything over 6:

$=(k+1)\left(\frac{k(2k+1)+6(k+1)}{6}\right).$

Simplify the numerator:

$k(2k+1)+6(k+1)=2k^2+k+6k+6=2k^2+7k+6.$

Factor the quadratic:

$2k^2+7k+6=(2k+3)(k+2).$

So

$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}.$

That is exactly the formula with $n=k+1$:

$\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}.$

Conclude the proof

Since the statement is true for $n=1$ and true for $n=k+1$ whenever it is true for $n=k$, the formula holds for all positive integers $n$ by mathematical induction.

Key idea to remember

Induction is not about checking many cases; it is about proving a chain: the base case starts the chain, and the inductive step shows that one true case forces the next one to be true. That is why the identity works for every positive integer.

💡 Pitfall guide

A common mistake is to substitute $k+1$ too early and lose the structure of the inductive hypothesis. You should first write the sum for $k+1$, then replace the first $k$ terms using the assumption. Another error is failing to factor correctly after combining fractions; the expression $2k^2+7k+6$ must become $(2k+3)(k+2)$, not some unrelated product. Finally, students sometimes forget that induction needs both the base case and the inductive step. Leaving out either one makes the proof incomplete.

🔄 Real-world variant

If the problem asked for the sum of cubes instead, such as $1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$, the induction structure would be similar but the algebra in the inductive step would be different. You would assume the cube-sum formula for $k$ and then show it for $k+1$ by adding $(k+1)^3$. The same proof strategy applies, but each identity has its own factorization pattern and base case.

🔍 Related terms

mathematical induction, inductive hypothesis, summation identity