\[S=\sum_{r=2}^{2014} \left[(-1)^r \dfrac{r^2+r+1}{r!}\right]\]

The value of \(S\) is of the form \[a+\dfrac{1}{b!}+\dfrac{1}{c!}\] Where \(a,b\) & \(c\) are integers and \(a\) & \(b\) are co-prime.

Find \[\dfrac{2a+2b+c}{4}\]

- Please avoid using Wolfram Alpha or any other CAS.

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