How can we find a new sum of multiplications based on a previous one? Suppose wehave two sequences:
$$(a_0, a_1, a_2, \dots, a_{2^n-1})$$
$$(b_0, b_1, b_2, \dots, b_{2^n-1})$$
We also have the following sum:
$$\sum_{k=0}^{2^n-1}{a_k \cdot b_k}$$
I'd like to know the fastest way to get this sum:
$$\sum_{k=0}^{2^n-1}{(a_k+c) \cdot (b_k+d)}$$
MY THOUGHTS AND WORK
I thought that some statiscal information could be crucial.  We can suppose that other statistics are handy - I will try to find a way to get them.  I'm mainly interested in the fastest way to calculate this problem.
I figured that if we could get some kind of geometric mean and perhaps standard deviation for both sequences, we might somehow be able to add $c$ and $d$ to these values and simply multiply them together.  I'm not very handy with statistics though, so I really don't know if this is possible.
WHAT I'D LIKE TO DO
I'm hoping that this could become a community wiki so that we could list all methods that might have a shot at being fast, and use feedback to find the best answer.
 A: Since it is a finite sum you can multiply out the brackets and split up the sums, to give
$$\sum_{k=0}^{2^n-1} (a_k+c)(b_k+d) = \sum_{k=0}^{2^n-1} a_kb_k + d\sum_{k=0}^{2^n-1} a_k + c\sum_{k=0}^{2^n-1} b_k + 2^ncd$$
So we calculate $\sum a_k$ (which is $2^n$ times the arithmetic mean of the $a_k$s) and $\sum b_k$ (analogous), and then it's just simple arithmetic.
But I'm not sure what you mean about finding the "fastest way". 

There is another perspective: you can define
$$ \begin{align} \mathbf{a} &= (a_0, a_2, \dots, a_{2^n-1}) \\ \mathbf{b} &= (b_0, b_2, \dots, b_{2^n-1}) \end{align}$$
Then we have
$$\sum_{k=0}^{2^n-1} a_kb_k = \mathbf{a} \cdot \mathbf{b}$$
If, further, we define $$\mathbf{1} = (\underbrace{1, 1, \dots, 1}_{2^n})$$
then we get
$$\sum_{k=0}^{2^n-1} (a_k+c)(b_k+d) = (\mathbf{a}+c\mathbf{1}) \cdot (\mathbf{b} + d\mathbf{1})$$
I'm not sure if this geometric picture helps, but it's something you could investigate.
A: You probably have three meaningful choices: work out each term of $$\sum_{k=0}^{2^n-1}{(a_k+c) \cdot (b_k+d)}$$ and then take the sum, or work out the four terms $$\sum_{k=0}^{2^n-1}{a_k  \cdot b_k}+d\sum_{k=0}^{2^n-1}{a_k}+c\sum_{k=0}^{2^n-1}{b_k}+c\cdot d \cdot 2^n ,$$ or the three terms $$\sum_{k=0}^{2^n-1}{a_k  \cdot b_k}+ \sum_{k=0}^{2^n-1}(d \cdot a_k+c \cdot b_k) +c\cdot d \cdot 2^n. $$  My money would be on the second. 
