I'm looking for the cumulative form of the following series:
$\sum_{i} (A+Bi)(2^i-C)$
A, B and C are all constants.
Not a mathematician by breed-- got the formula this far. Feels that it has something to do with binomial expansions, but can't put it together. Sorry if a naive question.
On
The distributive law is your friend, here.
$\begin{array}\\ \sum_{i=1}^n (A+Bi)(2^i-C) &=\sum_{i=1}^n(A2^i-AC+Bi2^i-BCi)\\ &=\sum_{i=1}^nA2^i-\sum_{i=1}^n AC+\sum_{i=1}^nBi2^i-\sum_{i=1}^n BCi\\ &=A\sum_{i=1}^n2^i-AC\sum_{i=1}^n 1+B\sum_{i=1}^ni2^i-BC\sum_{i=1}^n i\\ &=A\sum_{i=1}^n2^i-AC\sum_{i=1}^n 1+B\sum_{i=1}^ni2^i-BC\sum_{i=1}^n i\\ &=A(2^{n+1}-1)-ACn+B\sum_{i=1}^ni2^i-BCn(n+1)/2\\ \end{array} $
All the sums are easy except $s(n) =\sum_{i=1}^ni2^i $.
The more general formula is $\sum_{k=1}^n kx^k =\dfrac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2} $. Setting $x=2$, $\sum_{k=1}^n k2^k =n2^{n+2}-(n+1)2^{n+1}+2 $.
You can now evaluate your sum.
\begin{align}\sum_{i=1}^N(A+Bi)(2^i-C)&= \left(\sum_{i=1}^N 2^i(A+Bi)\right)-C\left(\sum_{i=1}^N(A+Bi) \right) \end{align}
Notice that $\left(\sum_{i=1}^N(A+Bi) \right)$ is just an arithmetic sum.
Now, to handle the first term: Let $$S=\sum_{i=1}^N 2^i(A+Bi)$$
$$2S=\sum_{i=1}^N 2^{i+1}(A+Bi)=\sum_{i=2}^{N+1}2^i(A+Bi-B)$$
$$S=2S-S=2^{N+1}(A+BN)-2(A+B)-B\sum_{i=2}^N2^i$$
Hence \begin{align}S&=2^{N+1}(A+BN)-2(A+B)-4B\sum_{i=0}^{N-2}2^i \\&=2^{N+1}(A+BN)-2(A+B)-4B(2^{N-1}-1)\\\end{align}