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

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
$$\left.\left(z\frac{d}{dz}\right)^2
(1+z)^n\right|_{z=1} =
\left.\left(z\frac{d}{dz}\right)
nz(1+z)^{n-1}\right|_{z=1}
\\= \left.zn(1+z)^{n-1} + n(n-1)z^2(1+z)^{n-2}\right|_{z=1}
= n2^{n-1} + n(n-1) 2^{n-2} 
\\= n(n+1) 2^{n-2}
= \sum_{q=0}^n q^2 {n\choose q}.$$
