How to calculate $\sum_{k=1}^n \left(k \sum_{i=0}^{k-1} {n \choose i}\right)$

Question

How do I calculate the following summation? $$\sum_{k=1}^n \left[k \sum_{i=0}^{k-1} {n \choose i}\right]$$

Answer 1

$\displaystyle\sum_{k=1}^n\left[\sum_{i=0}^{k-1} k{n\choose i}\right]=\sum_{i=0}^{n-1}\left[\sum_{k=i+1}^{n}k{n\choose i}\right]=\sum_{i=0}^{n-1}\left[{n\choose i}\sum_{k=i+1}^n k\right]=$

$\displaystyle = \sum_{i=0}^{n-1}{n\choose i} \left[\dfrac{n(n+1)}{2}-\dfrac{i(i+1)}{2}\right]=\dfrac{n(n+1)}{2}(2^{n}-1)-\sum_{i=0}^{n-1}\dfrac{i(i+1)}{2}{n\choose i}$

$\displaystyle =\dfrac{n(n+1)}{2}(2^{n}-1)-\sum_{i=0}^{n-1}\left[\dfrac{n(n-1)}{2}{n-2\choose i-2}+n{n-1\choose i-1}\right]$

$\displaystyle =\dfrac{n(n+1)}{2}(2^{n}-1)-\dfrac{n(n-1)}{2}(2^{n-2}-1)-n(2^{n-1}-1)$

$\displaystyle =n(3n+1).2^{n-3}$

Answer 2

We can do it by using generating functions:

Consider the g.f. for $\binom{n}{k}$ \begin{align*} (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \end{align*} and the sum of the coefficients can be given by $$\frac{(1+x)^n-2^n\, x^{n+1}}{1-x} = \sum_{k=0}^n \sum_{i=0}^k \binom{n}{i} x^k$$ Differentiating both sides w.r.t $x$ and taking $\displaystyle \lim_{x\to 1}$ gives $$\sum_{k=0}^n k\, \sum_{i=0}^k \binom{n}{i} = 2^{n - 3} {\left(3 \, n + 5\right)} n$$ Hence,

\begin{align*} \sum_{k=0}^n k\, \sum_{i=0}^{k-1} \binom{n}{i} &= \left(\sum_{k=0}^n k\, \sum_{i=0}^k \binom{n}{i}\right) - \left(\sum_{k=0}^n k\, \binom{n}{k}\right)\\\\ &=n {\left(3 \, n + 5\right)} \, 2^{n - 3} - n\, 2^{n-1}\\\\ &= n {\left(3 \, n + 1\right)} \, 2^{n - 3} \end{align*}

Answer 3

Hint 1: interchange the order of summation.

Hint 2: do you know how to find $\sum i {n \choose i}$?

Hint 3: do you know how to find $\sum i(i-1) {n \choose i}$?

Answer 4

(Note: I have attempted to incorporate OP's edit to the problem.)

$\begin{array}\\ \sum_{k=1}^n (k \sum_{i=0}^{k-1} {n \choose i}) &=\sum_{k=1}^n \sum_{i=0}^{k-1} (k {n \choose i})\\ &=\sum_{i=0}^{n-1} \sum_{k=i+1}^n (k {n \choose i})\\ &=\sum_{i=0}^{n-1} {n \choose i}\sum_{k=i+1}^n k \\ &=\sum_{i=0}^{n-1} {n \choose i}\left(\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\right) \\ &=\frac{n(n+1)}{2}\sum_{i=0}^{n-1} {n \choose i}-\sum_{i=0}^{n-1} {n \choose i}\frac{i(i+1)}{2} \\ &= S-T\\ \end{array} $

$S =\frac{n(n+1)}{2}\sum_{i=0}^{n-1} {n \choose i} =\frac{n(n+1)}{2}(\sum_{i=0}^{n} {n \choose i}-{n \choose n}) =\frac{n(n+1)}{2}(2^n-1) $.

$T = \sum_{i=0}^{n-1} {n \choose i}\frac{i(i+1)}{2}$.

Consider $t_i = {n \choose i}\frac{i(i+1)}{2}$.

$t_0 = 0, t_1=n$.

For $i \ge 1$, I would like to use the identity ${ n \choose m}{m \choose k} ={n \choose k}{n-k \choose m-k} $.

$\begin{array}\\ t_i &={n \choose i}\frac{i(i+1)}{2}\\ &={n \choose i}\left(\frac{i(i-1)}{2}+i\right)\\ &={n \choose i}({i \choose 2}+{i \choose 1})\\ &={n \choose i}{i \choose 2}+{n \choose i}{i \choose 1}\\ &={n \choose 2}{n-2 \choose i-2}+{n \choose 1}{n-1 \choose i-1}\\ \end{array} $

$\begin{array}\\ T &= t_0+t_1+\sum_{i=2}^{n-1} t_i\\ &= n+\sum_{i=2}^{n-1} \left( {n \choose 2}{n-2 \choose i-2}+{n \choose 1}{n-1 \choose i-1}\right)\\ &= n+\sum_{i=2}^{n-1} {n \choose 2}{n-2 \choose i-2}+\sum_{i=2}^{n-1}{n \choose 1}{n-1 \choose i-1}\\ &= n+ {n \choose 2}\sum_{i=2}^{n-1}{n-2 \choose i-2}+{n \choose 1}\sum_{i=2}^{n-1}{n-1 \choose i-1}\\ &= n+ {n \choose 2}\sum_{i=0}^{n-3}{n-2 \choose i}+{n \choose 1}\sum_{i=1}^{n-1}{n-1 \choose i}\\ &= n+ {n \choose 2}\left(\sum_{i=0}^{n-2}{n-2 \choose i}-1\right)+{n \choose 1}\sum_{i=0}^{n-1}{n-1 \choose i}-1\\ &= n+ {n \choose 2}\left(2^{n-2}-1\right)+{n \choose 1}(2^{n-1}-1)\\ &= n+ \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n(2^{n-1}-1)\\ &= \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n2^{n-1}\\ \end{array} $

$\begin{array}\\ S-T &=\frac{n(n+1)}{2}(2^n-1)-\left( \frac{n(n+1)}{ 2}\left(2^{n-2}-1\right)+n2^{n-1}\right)\\ &=2^{n-2}\left(2n(n+1)-\frac{n(n+1)}{ 2} -2n\right) -\frac{n(n+1)}{2}- \frac{n(n+1)}{ 2}\\ &=2^{n-2}\frac{3n(n+1)-4n}{ 2}-n(n+1)\\ &=2^{n-2}\frac{n(3n-1)}{ 2}-n(n+1)\\ \end{array} $

As usual, this was done off the top of my head, so there is a good chance of error, but it's late and I'm tired, so I'll stop here.

All corrections will be appreciated.

Answer 5

Here is a slightly different take on this that emphasises the use of formal power series.

Suppose we seek to evaluate $$Q_n = \sum_{k=1}^n k \sum_{q=0}^{k-1} {n\choose q}.$$

We have $$\sum_{q=0}^n {n\choose q} z^q = (1+z)^n$$ and therefore $$\sum_{q=0}^{k-1} {n\choose q} = [z^{k-1}] \frac{1}{1-z} (1+z)^n$$ because multiplication by $1/(1-z)$ sums coefficients.

The sum now becomes $$\sum_{k=1}^n k [z^{k-1}] \frac{1}{1-z} (1+z)^n = \sum_{k=0}^{n-1} (k+1) [z^k] \frac{1}{1-z} (1+z)^n$$

Once more deploying multiplication by $1/(1-z)$ this is equivalent to $$[z^{n-1}] \frac{1}{1-z} \sum_{k=0}^{n-1} z^k (k+1) [z^k] \frac{1}{1-z} (1+z)^n.$$

The operator sequence that extracts the coefficient on $z^k$, multiplies by $k+1$ and thereafter by $z^k$ is a generalized annihilated coefficient extractor and represents multiplication by $z$ followed by differentiation, so we get

$$Q_n = [z^{n-1}] \frac{1}{1-z} \left(\frac{z}{1-z} (1+z)^n\right)'.$$

Actually computing the derivative we obtain $$[z^{n-1}] \frac{1}{1-z} \left(\frac{1}{(1-z)^2} (1+z)^n + \frac{z}{1-z} \times n \times (1+z)^{n-1}\right)$$ which is $$[z^{n-1}] \frac{1}{(1-z)^3} (1+z)^n + n [z^{n-2}] \frac{1}{(1-z)^2} (1+z)^{n-1}.$$ This is $$\frac{1}{2} \sum_{q=0}^{n-1} {n\choose q} (n-1-q+1)(n-1-q+2) + n \sum_{q=0}^{n-2} {n-1\choose q} (n-2-q+1)$$ which is $$\frac{1}{2} \sum_{q=0}^{n-1} {n\choose q} (n-q)(n-q+1) + n \sum_{q=0}^{n-2} {n-1\choose q} (n-q-1) \\ = \frac{1}{2} \times n \times 2 + \frac{1}{2} \sum_{q=0}^{n-2} {n\choose q} (n-q)(n-q+1) + n \sum_{q=0}^{n-2} {n-1\choose q} (n-q-1)$$ which finally simplifies to $$n + \frac{1}{2} \sum_{q=0}^{n-2} {n\choose q} (n-q)(n-q+1) + \sum_{q=0}^{n-2} {n\choose q} (n-q-1) (n-q) \\ = n + \frac{3}{2} \sum_{q=0}^{n-2} {n\choose q} (n-q)^2 - \frac{1}{2} \sum_{q=0}^{n-2} {n\choose q} (n-q) \\ = n - \frac{3}{2} n + \frac{1}{2} n + \frac{3}{2} \sum_{q=0}^n {n\choose q} (n-q)^2 - \frac{1}{2} \sum_{q=0}^n {n\choose q} (n-q) \\ = \frac{3}{2} \sum_{q=0}^n {n\choose q} (n-q)^2 - \frac{1}{2} \sum_{q=0}^n {n\choose q} (n-q).$$

Recall the well-known identities (not difficult to prove) $$\sum_{q=0}^n q {n\choose q} = n 2^{n-1} \quad\text{and}\quad \sum_{q=0}^n q^2 {n\choose q} = n(n+1) 2^{n-2}$$ to finally obtain $$\frac{3}{2} n(n+1) 2^{n-2} -\frac{1}{2} n 2^{n-1} = n (3n+3) 2^{n-3} - 2n 2^{n-3} = n(3n+1) 2^{n-3}.$$

There is more on annihilated coefficient extractors at this MSE link.

Addendum. These last two identities can be proved with the operator given by $z\frac{d}{dz}.$ For the first identity we have $$\left.\left(z\frac{d}{dz}\right) (1+z)^n\right|_{z=1} = \left.nz(1+z)^{n-1}\right|_{z=1} = n2^{n-1} =\sum_{q=0}^n q {n\choose q}$$ and for the second one $$\left.\left(z\frac{d}{dz}\right)^2 (1+z)^n\right|_{z=1} = \left.\left(z\frac{d}{dz}\right) nz(1+z)^{n-1}\right|_{z=1} \\= \left.zn(1+z)^{n-1} + n(n-1)z^2(1+z)^{n-2}\right|_{z=1} = n2^{n-1} + n(n-1) 2^{n-2} \\= n(n+1) 2^{n-2} = \sum_{q=0}^n q^2 {n\choose q}.$$