SUM FUN

Motivations Recognise that this is a series type question (as in the tags), and one of the main ideas we should have in our mind would be to use telescope to eliminate most of the terms, leaving behind a manageable (or calculable) terms.

At first look, this question will seem daunting to most, so to start, with telescoping in mind, let's list out the first $2$ (counting backwards) general terms.

The $k^\text{th}$ term: $\dfrac{(k-1)k!}{2^k}$

The $(k-1)^\text{th}$ term: $\dfrac{(k-2)(k-1)!}{2^{k-1}}$ .

Here, we try to relate the $2$ terms in hopes of finding a telescoping series (using partial fractions). Since we want to relate the $2$ terms, we can try constructing a $2^{k-1}$ in the $k^\text{th}$ term: Let

$\dfrac{(k-1)k!}{2^k} = \dfrac{m}{2^k} + \dfrac{n}{2^{k-1}}$ .

Manipulating the above, we get: $m + 2n = (k-1)k! = (k+1-2)k! = (k+1)k! +- 2k!$

Therefore, $\dfrac{(k-1)k!}{2^k} = \dfrac{(k+1)!}{2^k} - \dfrac{k!}{2^{k-1}}$ .

Substituting this back into our equation, we get:

$\displaystyle\sum_{k=1}^{2014} \dfrac{(k-1)k!}{2^k} = \frac{2015!}{2^{2014}} - \frac{2014!}{2^{2013}} + \frac{2014!}{2^{2013}} - ... - \frac{2!}{2^1} + \frac{2!}{2^1} - \frac{1!}{2^0} = \frac{2015!}{2^{2014}} -1$

Therefore, $a = 2015, b= 2, c=2014$ and $a+b+c = \boxed{4031}$ .

$\begin{array}{cl} &\sum_{k=1}^{2014}\frac{(k-1)k!}{2^k}\\ =&\sum_{k=1}^{2014}\frac{(k+1-2)k!}{2^k}\\ =&\sum_{k=1}^{2014}\frac{(k+1)k!-2k!}{2^k}\\ =&\sum_{k=1}^{2014}\frac{(k+1)!-2k!}{2^k}\\ =&\sum_{k=1}^{2014}\frac{(k+1)!}{2^k}-\sum_{k=1}^{2014}\frac{2k!}{2^k}\\ =&\sum_{k=1}^{2014}\frac{(k+1)!}{2^k}-\sum_{k=1}^{2014}\frac{k!}{2^{k-1}}\\ =&\sum_{k=1}^{2014}\frac{(k+1)!}{2^k}-\sum_{k=0}^{2013}\frac{(k+1)!}{2^k}\\ =&\left.\frac{(k+1)!}{2^k}\right|_{x=2014}-\left.\frac{(k+1)!}{2^k}\right|_{x=0}\\ =&\frac{2015!}{2^{2014}}-1 \end{array}$

$\displaystyle \sum_{k=1}^{2014} \frac{(k-1)k!}{2^{k}}$ $=\displaystyle \sum_{k=1}^{2014} (\frac{(k+1-1)k!}{2^{k}} - \frac{k!}{2^{k}})$ $=\displaystyle \sum_{k=1}^{2014} (\frac{(k+1)!}{2^{k}} - \frac{k!}{2^{k-1}}) = \frac{2015!}{2^{2014}} - 1$