Increasing Sum of Binomial Coefficients

Theorem

Let $n \in \Z$ be an integer.

Then:

$\ds \sum_{j \mathop = 0}^n j \binom n j = n 2^{n - 1}$

where $\dbinom n k$ denotes a binomial coefficient.

That is:

$1 \dbinom n 1 + 2 \dbinom n 2 + 3 \dbinom n 3 + \dotsb + n \dbinom n n = n 2^{n - 1}$

Proof 1

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

$=$

$\ds \sum_{j \mathop = 1}^n j \binom n j$

as $0 \dbinom n 0 = 0$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^n n \binom {n - 1} {j - 1}$

Factors of Binomial Coefficient

$\ds $

$=$

$\ds n \sum_{j \mathop = 0}^{n - 1} \binom {n - 1} j$

Translation of Index Variable of Summation

$\ds $

$=$

$\ds n 2^{n - 1}$

Sum of Binomial Coefficients over Lower Index

$\blacksquare$

Proof 2

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.

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion.
In particular: $2$ further approaches are offered by 1980: David M. Burton: Elementary Number Theory (revised ed.) as hints in the exercise in which this appears
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Sources

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.14$
1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(c)}$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.14.$