Telescopic 2014

Let us seperate the original equation

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

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

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

Seperating(r -1 + 1), then we will solve the 2 equations seperately by $V_n$ method

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

All the terms will cancel out as put the values of 'r' except first and last term

We will now get

$\displaystyle (1 + \frac{1}{2013!}) + (1 + \frac{1}{2014!})$

you should specify about b and c as answer may change

Here

a = 2

b = 2013

c = 2014

$\displaystyle \frac{2a + 2b + c}{4} = \boxed{1511}$

$\displaystyle \sum_{r=2}^{2014} (-1)^r \bigg(\dfrac{r^2+r+1}{r!}\bigg)$

= $\displaystyle \sum_{r=2}^{2014} (-1)^r \bigg(\dfrac{r}{(r-1)!} + \dfrac{r+1}{r!}\bigg)$

= $\displaystyle \sum_{r=2}^{2014} \bigg( (-1)^r \dfrac{r}{(r-1)!} - (-1)^{r+1} \dfrac{r+1}{(r+1)!}\bigg)$

Now , this is clearly a telescoping series, which telescopes to

$2 + \dfrac{2015}{2014!} = 2+\dfrac{1}{2013!} + \dfrac{1}{2013!}$

Clearly, $a=2$ , $b=2013$ , $c = 2014$