Easy Telescoping Factorial Sum

Algebra Level 3

x = 101 ! × ( 1 2 ! + 2 3 ! + 3 4 ! + ⋯ + 99 100 ! ) x = 101! \times \left(\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} +\cdots+ \frac{99}{100!}\right)

Find 101 ! − x . 101! - x .


The answer is 101.

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Aayush Gupta by Aayush Gupta , 21, USA

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1 solution

If, instead, x = 101 ! ∗ ( 1 2 ! + 2 3 ! + 3 4 ! + . . . + 99 100 ! ) = 101 ! ∗ A x = 101! * (\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{99}{100!}) = 101! * A , then we can proceed as follows.

Note that n ( n + 1 ) ! = ( n + 1 ) − 1 ( n + 1 ) ! = 1 n ! − 1 ( n + 1 ) ! \frac{n}{(n+1)!} = \frac{(n+1) - 1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!} . Then

A = ( 1 1 ! − 1 2 ! ) + ( 1 2 ! − 1 3 ! ) + ( 1 3 ! − 1 4 ! ) + . . . . . + ( 1 98 ! − 1 99 ! ) + ( 1 99 ! − 1 100 ! ) = 1 − 1 100 ! A = (\frac{1}{1!} - \frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} - \frac{1}{4!}) + ..... + (\frac{1}{98!} - \frac{1}{99!}) + (\frac{1}{99!} - \frac{1}{100!}) = 1 - \frac{1}{100!} ,

since all the 'middle' terms cancel pairwise, (hence the reference to "telescoping" in the title). So we now have

101 ! − x = 101 ! − 101 ! ∗ ( 1 − 1 100 ! ) = 101 ! 100 ! = 101 101! - x = 101! - 101!*(1 - \frac{1}{100!}) = \frac{101!}{100!} = \boxed{101} .

18 Helpful 0 Interesting 0 Brilliant 1 Confused

Great solution! Really clear and concise :)

John Frank - 5 years, 3 months ago

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