Sum of Sequence of Squares/Proof by Binomial Coefficients

Theorem

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

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:

\(\ds 2 \binom n 2 + \binom n 1\)

\(=\)

\(\ds 2 \dfrac {n \paren {n - 1} } 2 + n\)

\(\ds \)

\(=\)

\(\ds n \paren {n - 1} + n\)

\(\ds \)

\(=\)

\(\ds n^2\)

Hence:

\(\ds \sum_{i \mathop = 1}^n i^2\)

\(=\)

\(\ds \sum_{i \mathop = 1}^n \paren {2 \binom i 2 + \binom i 1}\)

\(\ds \)

\(=\)

\(\ds 2 \binom {n + 1} 3 + \binom {n + 1} 2\)

Sum of Binomial Coefficients over Upper Index

\(\ds \)

\(=\)

\(\ds \frac {2 \paren {n + 1} n \paren {n - 1} } {3 \times 2 \times 1} + \frac {\paren {n + 1} n} 2\)

Definition of Binomial Coefficient

\(\ds \)

\(=\)

\(\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6\)

after algebra

$\blacksquare$

Also see

• Sum of Sequence of Fourth Powers/Proof using Binomial Coefficients

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $4 \ \text {(b)}$
• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients