Summed Fraction Factorials

Algebra Level 5

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + 5 3 ! + 4 ! + 5 ! + … + 2001 1999 ! + 2000 ! + 2001 ! = a b − c d ! \frac {3}{1! + 2! + 3!} + \frac {4}{2! + 3! + 4!} + \frac {5}{3! + 4! + 5!} + \ldots + \frac {2001}{1999!+ 2000! + 2001!} = \frac {a}{b} - \frac {c}{d!}

The equation above holds true for coprime postive integers a a and b b , and c c and d d . What is the digit sum of ( a b c d ) ( a b c ) ( a b ) a (abcd)^{(abc)^{(ab)^a}} ?


The answer is 54.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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2 solutions

Benjamin Tong
Apr 4, 2014

∑ n = 3 2001 n ( n − 2 ) ! + ( n − 1 ) ! + n ! = \sum_{n=3}^{2001} \frac{n}{(n-2)!+(n-1)!+n!} =

∑ n = 3 2001 n ( n − 2 ) ! ( 2 n ) = \sum_{n=3}^{2001}\frac{n}{(n-2)!(2n)} =

∑ n = 3 2001 1 n ( n − 2 ) ! = \sum_{n=3}^{2001} \frac{1}{n(n-2)!} =

∑ n = 3 2001 n − 1 n ! = \sum_{n=3}^{2001} \frac{n-1}{n!} =

∑ n = 3 2001 ( 1 ( n − 1 ) ! − 1 n ! ) = \sum_{n=3}^{2001} (\frac{1}{(n-1)!} - \frac{1}{n!}) =

1 2 − 1 2001 ! \frac{1}{2} - \frac{1}{2001!}

Thus what we want to calculate is 400 2 4 = 256512384128016 4002^4 = 256512384128016

And the digit sum is 54 \boxed{54}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

I believe there is a typo in the second line.

It should read (n-2)! n^2 in the denominator instead of (n-2)! (2n).

Alex Wang - 6 years, 7 months ago
Bony Roy
Feb 28, 2014

Simplify the expression to give 1 2 − 1 2001 ! \frac { 1 }{ 2 } -\frac { 1 }{ 2001! } abcd=4002, abc=2, ab=2, a=1 Calculate sum of digits of 4002 4 { 4002 }^{ 4 } . (I used an online calculator for this part) This gives 256512384128016. Digit Sum is 54

1 Helpful 0 Interesting 0 Brilliant 0 Confused

how to simplify the expression in the form as mentioned by you?

Sanjeev Prasad - 7 years, 3 months ago

I added the first few patterns and noticed the pattern, that the sum is 1 2 − 1 n ! \frac { 1 }{ 2 } -\frac { 1 }{ n! } where n is the numerator of the last term in the series. I have proved it by induction, but the proof is quite long so it would take a while to post it here. But essentially you are trying to prove that 1 2 − ( 1 k ! − k + 1 ( k − 1 ) ! + k ! + ( k + 1 ) ! ) = 1 2 − 1 ( k + 1 ) ! \frac { 1 }{ 2 } -\left( \frac { 1 }{ k! } -\frac { k+1 }{ (k-1)!+k!+(k+1)! } \right) =\frac { 1 }{ 2 } -\frac { 1 }{ (k+1)! }

Bony Roy - 7 years, 3 months ago

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