Factorial!

Number Theory Level 4

If the expression 3 1 ! + 2 ! + 3 ! \frac{3}{1!+2!+3!} + 4 2 ! + 3 ! + 4 ! \frac{4}{2!+3!+4!} + ...... + 2001 1999 ! + 2000 ! + 2001 ! \frac{2001}{1999!+2000!+2001!}

Can be written as 1 a ! \frac{1}{a!} - 1 b ! \frac{1}{b!} , Find a+b.


The answer is 2003.

Sean Ty by Sean Ty , 21, Philippines

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1 solution

Note first that

n! + (n+1)! + (n+2)! = n! * (1 + (n+1) + (n+1)(n+2)) = n! * (n+2)^2.

Thus (n+2) / (n! + (n+1)! + (n+2)!) = 1 / (n! * (n+2)) = (n+1) / (n+2)! =

[(n+2) / (n+2)!] - [1 / (n+2)!] = [1 / (n+1)!] - [1 / (n+2)!].

So now we end up with a telescoping series with n going from 1 to 1999. All the terms will cancel except for the first and last, leaving us with [1 / 2!] - [1 / 2001!], making a=2 and b=2001, and so a + b = 2003 is the final answer.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

nice solution! almost did the same as you!!

Kartik Sharma - 6 years, 10 months ago

overrated...

math man - 6 years, 8 months ago

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