The sum of the squares of the first \( n \) natural numbers is given by: \( \displaystyle S_n = \frac{n(n + 1)(2n + 1)}{6} \). In this post, we will learn how to prove it.
Proof:
We will prove it using mathematical induction:
Base Case \( ( n = 1 ) \):
\( \displaystyle S_1 = 1^2 = 1 = \frac{1(1 + 1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \)
The formula holds for \( n = 1 \).
Inductive Step:
Assume the formula holds for \( n = k \), i.e.,
\( \displaystyle S_k = 1^2 + 2^2 + \dots + k^2 = \frac{k(k + 1)(2k + 1)}{6} \)
Prove for \( n = k + 1 \):
\( \begin{aligned} &S_{k+1} = S_k + (k + 1)^2 = \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 \\ &= \frac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} \\ &= \frac{(k + 1)\left[k(2k + 1) + 6(k + 1)\right]}{6} \\ &= \frac{(k + 1)(2k^2 + k + 6k + 6)}{6} \\ &= \frac{(k + 1)(2k^2 + 7k + 6)}{6} \\ &= \frac{(k + 1)(k + 2)(2k + 3)}{6} \\ &= \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} \end{aligned} \)Thus, the formula holds for \( n = k + 1 \).
By induction, the formula is true for all \( n \in \mathbb{N} \).
Please let me know in the comments if you find any errors in this proof.