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Compute \(\dfrac{\big(100! + 99!\big) \times \big(98! + 97!\big) \times \ldots \big(2! + 1!\big)}{\big(100! - 99!\big) \times \big(98! - 97!\big) \times \ldots \big(2! - 1!\big)}\)

 Jun 9, 2020
 #1
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First, you have to sum them separately using "closed form" formulas:


The top one:= product_(n=1)^50((101 - 2 n)! + (-2 (-51 + n))!) =7.733930684 E+3510


The bottom one: = product_(n=1)^50(-(101 - 2 n)! + (-2 (-51 + n))!) = 7.657357113 E+3508


7.733930684 E+3510 / 7.657357113 E+3508 =101 - The result

 Jun 9, 2020
 #2
avatar+21835 
+1

Another way:   (doing a lot of factoring)

 

Numerator:  100! + 99!  =  99!(100 + 1)  =  99! · 101

                        98! + 97!  =  97!(98 + 1)  =  97! · 99

                        96! + 95!  =  95!(96 + 1)  =  95! · 97

                                  ...

                                2! + 1!  =  1!(2 + 1)  =  1! · 3

 

Total numerator:  99! · 101 · 97! · 99 · 95! · 97 · ... · 1! ·3

 

Denominator:  100! - 99!  =  99!(100 - 1)  =  99! · 99

                            98! - 97!  =  97!(98 - 1)  =  97! · 97

                            96! - 95!  =  95!(96 - 1)  =  95! · 95

                                     ...

                                  2! - 1!  =  1!(2 - 1)  =  1! · 1

 

Total denominator:  99! · 99 · 97! · 97 · 95! · 95 · 1! · 1

 

Notice that all the factorials of the numerator and the denominator cancel each other.

Notice that all the single numbers cancel each other except for the 101 in the numerator

and th 1 in the denominator.

 

Answer  =  101 / 1  =  101

 Jun 10, 2020

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