Σk² = n(n+1)(2n+1)/6. Where does this formula come from? Derivation and application.

Find Sum of Series: 1² + 2² + 3² + ... + n² — Special Sum Formula

Question

Find the sum of the series $1^2 + 2^2 + 3^2 + \cdots + n^2$ .

In other words, prove that:

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

This formula carries guaranteed weightage in CBSE Class 11 board exams and appears as a stepping stone in JEE problems involving series manipulation.

Solution — Step by Step

We want a clever identity that brings $k^2$ into play. Use the cubic expansion:

(k+1)^3 - k^3 = 3k^2 + 3k + 1

Why this particular identity? Because when we sum both sides over $k = 1$ to $n$ , the left side telescopes — almost everything cancels. That’s the trick.

\sum_{k=1}^{n} \left[(k+1)^3 - k^3\right] = \sum_{k=1}^{n} (3k^2 + 3k + 1)

The left side telescopes:

(2^3 - 1^3) + (3^3 - 2^3) + \cdots + ((n+1)^3 - n^3) = (n+1)^3 - 1

So we get:

(n+1)^3 - 1 = 3\sum k^2 + 3\sum k + \sum 1

We already know $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\displaystyle\sum_{k=1}^{n} 1 = n$ .

Substituting:

(n+1)^3 - 1 = 3S_2 + 3 \cdot \frac{n(n+1)}{2} + n

where $S_2 = \sum k^2$ is what we want.

Expand $(n+1)^3 = n^3 + 3n^2 + 3n + 1$ , so $(n+1)^3 - 1 = n^3 + 3n^2 + 3n$ .

Now rearrange:

3S_2 = n^3 + 3n^2 + 3n - \frac{3n(n+1)}{2} - n

3S_2 = n^3 + 3n^2 + 2n - \frac{3n(n+1)}{2}

Factor $n$ from the first three terms:

3S_2 = n(n^2 + 3n + 2) - \frac{3n(n+1)}{2} = n(n+1)(n+2) - \frac{3n(n+1)}{2}

Take $n(n+1)$ common:

3S_2 = n(n+1)\left[(n+2) - \frac{3}{2}\right] = n(n+1) \cdot \frac{2n+4-3}{2} = n(n+1) \cdot \frac{2n+1}{2}

Divide both sides by 3:

\boxed{S_2 = \frac{n(n+1)(2n+1)}{6}}

Why This Works

The telescoping method is the standard derivation you’ll see in NCERT and most board solutions. The key insight is that $k^3$ differences give us $k^2$ terms — we’re essentially “one degree up” to get what we want.

This mirrors how $\sum k$ is derived using $(k+1)^2 - k^2 = 2k + 1$ (one degree up from $k$ ). So for $\sum k^3$ , we’d use $(k+1)^4 - k^4$ . Same pattern, each time.

\sum_{k=1}^{n} 1 = n

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2

Alternative Method: Induction

If derivation isn’t asked and you just need to verify the formula (common in JEE objective questions), Mathematical Induction is faster.

Base case: $n = 1$ . LHS $= 1^2 = 1$ . RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$ . ✓

Inductive step: Assume true for $n = m$ , i.e., $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$ .

For $n = m+1$ :

\sum_{k=1}^{m+1} k^2 = \frac{m(m+1)(2m+1)}{6} + (m+1)^2

= (m+1)\left[\frac{m(2m+1)}{6} + (m+1)\right] = (m+1) \cdot \frac{2m^2 + 7m + 6}{6}

= \frac{(m+1)(m+2)(2m+3)}{6}

This is exactly $\frac{(m+1)((m+1)+1)(2(m+1)+1)}{6}$ . Induction complete. ✓

Induction is cleaner when you already know the formula — use it in objective settings to save time.

Common Mistake

Plugging in wrong value of $n$ . Students often compute $1^2 + 2^2 + \cdots + 10^2$ and use the formula with $n = 10$ , but then make the arithmetic error $\frac{10 \cdot 11 \cdot 21}{6}$ incorrectly as $\frac{10 \cdot 11 \cdot 20}{6}$ — forgetting the $(2n+1)$ term uses $2(10)+1 = 21$ , not $2n = 20$ .

Always write out all three factors explicitly: $n$ , $(n+1)$ , $(2n+1)$ — then substitute. Don’t rush the substitution.

Quick sanity check: For $n = 3$ , we get $1 + 4 + 9 = 14$ . Formula gives $\frac{3 \cdot 4 \cdot 7}{6} = 14$ . If your formula gives a non-integer or something obviously wrong for small $n$ , you’ve made an error somewhere.

Question

Solution — Step by Step

Why This Works

Alternative Method: Induction

Common Mistake

Try These Next