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.

$\blacksquare$


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