Fearsome Factorials!

Algebra Level 3

If the value of

∑ k = 1 2014 k ( k + 1 ) ! \sum^{2014}_{k =1} \frac{k}{(k + 1)!}

can be written in the form of

A + ( − 1 ) B ⋅ 1 C ! A + (-1)^{B} \cdot \frac{1}{C!} , when A , B A, B and C C are postive integers and B B is either 1 1 or 2 2 ,

find the value of A + B + C A + B + C .


The answer is 2017.

Sanchayapol Lewgasamsarn by Sanchayapol Lewgasamsarn , 20, Thailand

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1 solution

∑ k = 1 2014 k ( k + 1 ) ! \sum^{2014}_{k =1} \frac{k}{(k + 1)!}

= ∑ k = 1 2014 ( k + 1 ) − 1 ( k + 1 ) ! = \sum^{2014}_{k =1} \frac{(k+1) - 1}{(k + 1)!}

= ∑ k = 1 2014 1 k ! − 1 ( k + 1 ) ! = \sum^{2014}_{k =1} \frac{1}{k!} - \frac{1}{(k + 1)!}

= ( 1 1 ! − 1 2 ! ) + ( 1 2 ! − 1 3 ! ) + . . . + ( 1 2014 ! − 1 2015 ! ) = (\frac{1}{1!} - \frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + ... + (\frac{1}{2014!} - \frac{1}{2015!})

The above line is a Telescoping Series

= 1 − 1 2015 ! = 1 - \frac{1}{2015!}

= 1 + ( − 1 ) 1 ⋅ 1 2015 ! = 1 +(-1)^{1} \cdot \frac{1}{2015!}

Thus A = 1 , B = 1 , C = 2015 A =1, B =1, C = 2015

Thus A + B + C = 2017 A + B + C = \boxed{2017}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Really great

suraj prajapat - 7 years, 1 month ago

Creativity at its best

Star Light - 7 years, 1 month ago

Awesome problem, Sanchayapol. Initially, yes it was fearsome...but once you get the logic its just a matter of few seconds. I look forward to more such great problems from you :)

Krishna Ar - 7 years ago

Nice problem! Did the same way as you did! It is a good result even! Used in many problems on Brilliant even, I think...

Kartik Sharma - 6 years, 10 months ago

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