Sum of Sequence of Squares/Proof by Binomial Coefficients

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Theorem

$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$


Proof

From Binomial Coefficient with One:

$\dbinom n 1 = n$

From Binomial Coefficient with Two:

$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$


Thus:

\(\displaystyle 2 \binom n 2 + \binom n 1\) \(=\) \(\displaystyle 2 \dfrac {n \paren {n - 1} } 2 + n\)
\(\displaystyle \) \(=\) \(\displaystyle n \paren {n - 1} + n\)
\(\displaystyle \) \(=\) \(\displaystyle n^2\)

Hence:

\(\displaystyle \sum_{i \mathop = 1}^n i^2\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \paren {2 \binom i 2 + \binom i 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \binom {n + 1} 3 + \binom {n + 1} 2\) Sum of Binomial Coefficients over Upper Index
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \paren {n + 1} n \paren {n - 1} } {3 \times 2 \times 1} + \frac {\paren {n + 1} n} 2\) Definition of Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \frac {n \paren {n + 1} \paren {2 n + 1} } 6\) after algebra

$\blacksquare$


Sources