Not very hard but not very easy

Algebra Level 2

Evaluate ∑ n = 1 100 n ! × n \sum\limits_{n=1}^{100} n!\times n . The solution will be of the form a ! − b a!-b , where a a and b b are positive integers. Give your answer as a + b a+b .


The answer is 102.

Magnus Seidenfaden by Magnus Seidenfaden , 19, Denmark

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1 solution

Hugh Sir
Feb 3, 2019

n ! ∗ n = ( n ! ) ( n + 1 ) − n ! = ( n + 1 ) ! − n ! n!*n = (n!)(n+1) - n! = (n+1)! - n!

∑ n = 1 100 n ! ∗ n = ∑ n = 1 100 [ ( n + 1 ) ! − n ! ] = ∑ n = 1 100 ( n + 1 ) ! − ∑ n = 1 100 n ! = 101 ! − 1 ! = 101 ! − 1 = a ! − b \sum\limits_{n=1}^{100} n!*n = \sum\limits_{n=1}^{100} [(n+1)! - n!] = \sum\limits_{n=1}^{100} (n+1)! - \sum\limits_{n=1}^{100} n! = 101! - 1! = 101! - 1 = a! - b

So a = 101 a = 101 and b = 1 b = 1 .

Therefore, a + b = 102 a+b = 102 .

1 Helpful 0 Interesting 3 Brilliant 0 Confused

How did you evaluate the summations in the penultimate step?

Abhinav Raja - 2 years, 4 months ago

∑ n = 1 100 ( n + 1 ) ! − ∑ n = 1 100 n ! = ( 2 ! + 3 ! + 4 ! + . . . + 99 ! + 100 ! + 101 ! ) − ( 1 ! + 2 ! + 3 ! + . . . + 98 ! + 99 ! + 100 ! ) = 101 ! − 1 ! \sum\limits_{n=1}^{100} (n+1)! - \sum\limits_{n=1}^{100} n! = (2!+3!+4!+...+99!+100!+101!) - (1!+2!+3!+...+98!+99!+100!) = 101! - 1!

Hugh Sir - 2 years, 4 months ago

a = 102 , b = 102 ! − 101 ! + 1 a=102, b=102!-101!+1 . I think you need to say a a is as small as possible

X X - 2 years, 4 months ago

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