Increasing Sum of Binomial Coefficients/Proof 2
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Theorem
- $\ds \sum_{j \mathop = 0}^n j \binom n j = n 2^{n - 1}$
Proof
From the Binomial Theorem:
- $(1): \quad \paren {1 + x}^n = \ds \sum_{j \mathop = 0}^n \dbinom n j x^n$
Differentiating $(1)$ with respect to $x$:
\(\ds n \paren {1 + x}^{n - 1}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^n j \dbinom n j x^{j - 1}\) | Power Rule for Derivatives | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds n \paren {1 + 1}^{n - 1}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^n j \dbinom n j\) | setting $x = 1$ |
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem