Does $ \sum_{k=0}^n k {n \choose k}$ have a convenient exponential equivalent?

Question

I know that: $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$

Does $$0 {n \choose 0} + 1 {n \choose 1} + ... + n {n \choose n} = ??$$

have some convenient simplification?

Answer 1

Using $$\displaystyle (1+x)^n = \binom{n}{0}x^0+\binom{n}{1}x+\binom{n}{2}x^2+.......+\binom{n}{n}x^n$$

Now Differentiate both side w. r to $x\;,$ We get

$$\displaystyle n(1+x)^{n-1} = 0+1\cdot \binom{n}{1}x^{0}+2\cdot \binom{n}{2}x^{1}+..........+n\cdot \binom{n}{n}x^{n-1}$$

Now Put $x=1\;,$ we get

$$\displaystyle n\cdot 2^{n-1} = \binom{n}{1}+2\cdot \binom{n}{2}+.........+n\cdot \binom{n}{n}$$

Answer 2

$$\sum\limits_{k=1} ^n k\cdot \binom{n}{k} = 1 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + 3 \cdot \binom{n}{3} + \ldots + n \cdot \binom{n}{n} = n \cdot 2^{n-1}$$ valid $\forall n \in \mathbb{N}$.

Proof $$\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}$$ $$=n\binom{n-1}{0}+2\frac{n}{2}\binom{n-1}{1}+3\frac{n}{3}\binom{n-1}{2}+\cdots+n\binom{n-1}{n-1}$$ $$=n\binom{n-1}{0}+n\binom{n-1}{1}+n\binom{n-1}{2}+\cdots+n\binom{n-1}{n-1}$$ $$=n\left(\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\cdots+\binom{n-1}{n-1}\right)=n\cdot 2^{n-1}$$

Answer 3

For such a problem, it is always convenient to look at the underlying power series. Recall that $$\sum_{k=0}^{\infty} \binom{n}{k}x^k=(x+1)^n$$ which is called the Binomial Theorem. The special case $x=1$ was mentioned above.

Now, if we take the derivative on both sides, we obtain $$\sum_{k=1}^{\infty} k \cdot \binom{n}{k} x^{k-1}=n(x+1)^{n-1}$$ and hence in the special case $x=1$ your sum evaluates to $n2^{n-1}$.

Remark: Note that in this case we don't have to examine tedious things such as the radius of convergence since the non-zero part of all the series terminates by the definition of the binomial coefficients.