Sum of Binomial Coefficients over Lower Index/Proof 3

Theorem

$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$

Proof

From the Binomial Theorem, we have that:

$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$

Putting $x = y = 1$ we get:

\(\ds 2^n\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 0}^n \binom n i 1^{n - i} 1^i\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 0}^n \binom n i\)

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients
• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(a)}$