Compute \(\dfrac{\big(100! + 99!\big) \times \big(98! + 97!\big) \times \ldots \big(2! + 1!\big)}{\big(100! - 99!\big) \times \big(98! - 97!\big) \times \ldots \big(2! - 1!\big)}\)
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
+23246 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
