Sum of the SQUARES of the first $n$ natural numbers, $1^2 + 2^2 + 3^2 …

Question

Sum of the SQUARES of the first $n$ natural numbers, $1^2 + 2^2 + 3^2 + \ldots + n^2$, equals

Accepted Answer

C. $n(n+1)(2n+1)/6$

1. NCERT §8.5 (Sum of squares formula): $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$.
2. Examples: $n = 3$: $1 + 4 + 9 = 14$; formula gives $3 \cdot 4 \cdot 7 / 6 = 84/6 = 14$ ✓.
3. $n = 10$: formula gives $10 \cdot 11 \cdot 21 / 6 = 2310/6 = 385$ ✓.
4. Option A is the sum of FIRST POWERS (linear). Option B is approximate but wrong. Option D is the sum of CUBES formula.
_Source: NCERT Class 11 Mathematics, Ch 8, §8.5 (Sum of squares — Eq. 8.27), p. 13._

Answer

$n(n+1)/2$

Answer

$n^3/3$

Answer

$(n(n+1)/2)^2$