Find sum of first n terms of 1² + 2² + 3² + ... + n²

Question

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

Solution — Step by Step

We claim:

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

This is one of three standard sum formulas that every student should memorise. Let’s derive it.

Use the identity: $k^3 - (k-1)^3 = 3k^2 - 3k + 1$

Write this for $k = 1, 2, 3, \ldots, n$ and sum:

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

Left side (telescoping — most terms cancel):

n^3 - 0^3 = n^3

Right side:

3\sum_{k=1}^{n} k^2 - 3\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1

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

where $S = \sum_{k=1}^n k^2$ (what we want to find) and we used the known formula $\sum k = \frac{n(n+1)}{2}$ .

Setting left = right:

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

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

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

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

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

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

For $n = 3$ : $1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$ .

Formula: $\frac{3 \times 4 \times 7}{6} = \frac{84}{6} = 14$ ✓

For $n = 4$ : $1+4+9+16 = 30$ .

Formula: $\frac{4 \times 5 \times 9}{6} = \frac{180}{6} = 30$ ✓

Why This Works

The telescoping identity $k^3 - (k-1)^3$ is chosen specifically because it contains $k^2$ — allowing us to isolate the sum of squares once we know the sum of $k$ and the constant term. This method generalises: to find $\sum k^3$ , use the identity $(k^4 - (k-1)^4)$ and so on.

Alternatively, this result can be proved by mathematical induction:

Base case: $n=1$ : $1 = \frac{1 \times 2 \times 3}{6} = 1$ ✓
Inductive step: Assume true for $n=k$ ; add $(k+1)^2$ to both sides and simplify to get the formula with $n=k+1$ .

Alternative Method — Proof by Induction (Outline)

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

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

Inductive step: Assume true for $n=m$ . 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{m(2m+1) + 6(m+1)}{6}

= (m+1) \cdot \frac{2m^2+7m+6}{6} = (m+1) \cdot \frac{(m+2)(2m+3)}{6} = \frac{(m+1)(m+2)(2m+3)}{6}

This is the formula with $n = m+1$ : $\frac{(m+1)((m+1)+1)(2(m+1)+1)}{6}$ ✓

\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

All three formulas are tested in CBSE Class 11 and JEE Main. Memorise all three. For $\sum k^3$ , note the beautiful result: it equals $(\sum k)^2$ — the cube sum equals the square of the simple sum.

Common Mistake

Students often write the formula as $\frac{n(n+1)(2n+1)}{3}$ (dividing by 3 instead of 6). The correct denominator is 6. Verify: for $n=1$ , the formula must give 1. With denominator 3: $\frac{1 \times 2 \times 3}{3} = 2 \neq 1$ . With denominator 6: $\frac{1 \times 2 \times 3}{6} = 1$ ✓. Always verify with $n=1$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Proof by Induction (Outline)

Common Mistake

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