( x + 1 ) ( x + 2 ) ⋯ ( x + 9 9 9 ) 1
The partial fraction decomposition of the expression above can be expressed as m = 1 ∑ 9 9 9 x + m a m for some a 1 , a 2 , … , a 9 9 9 .
If max { ∣ a 1 ∣ , ∣ a 2 ∣ , … , ∣ a 9 9 9 ∣ } = ( Q ! ) 2 1 , find Q .
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The cover-up rule is an underutilized technique for partial fraction decomposition. Note that the solution manages to explain how to find the value of ALL the values of a j in one single step!
Since the coefficient a j 1 is in the form of a ! b ! , where a + b is a constant, it is a natural to consider rewriting the value of a j 1 in terms of the binomial coefficient ( a a + b ) = ( b a + b ) = a ! b ! ( a + b ) ! .
OH! So this is called the cover-up rule. My teacher used to call it the "takip" method (literally covering method) when he taught us this.
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It's an underrated technique!!
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Agree. I only saw it for two times in my entire math career: Once from my teacher, twice here.
GREAT PROBLEM BUT I HAD SOLVED SIMILAR PROBLEMS EARLIER SO NOT A DIFFICULT ONE
OKAY! THANK YOU! WHY ARE YOU TYPING IN CAPS LOCK????!?!?!?!?!?!?!?!
Ahh... SORRY Buddy
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Relevant wiki: Partial Fractions - Cover Up Rule
Naturally, the first step is to find the closed form of a j where j = 1 , 2 , … , 9 9 9 .
By cover up rule , the value of a j 1 can be expressed as
= ± ( j − 1 ) ! ( − j + 1 ) ( − j + 2 ) ⋯ ( − j + j − 2 ) ( − j + j − 1 ) ( x + j ) 1 = ( 9 9 9 − j ) ! ( − j + j + 1 ) ( − j + j + 2 ) ⋯ ( − j + 9 9 9 ) = ± ( j − 1 ) ! ( 9 9 9 − j ) !
Hence, ∣ a j ∣ = ∣ ∣ ∣ ∣ ( j − 1 ) ! ( 9 9 9 − j ) ! 1 ∣ ∣ ∣ ∣ ⇔ ∣ a j ∣ ⋅ 9 9 8 ! = ( j − 1 ) ! ( 9 9 9 − j ) ! ( ( j − 1 ) + ( 9 9 9 − j ) ) ! = ( j − 1 9 9 8 ) .
This leaves us with finding the maximum (integer) value of ( j − 1 9 9 8 ) .
By looking at the Pascal's triangle , the largest value of ( A 2 N ) for constant N > 0 occurs when A = N . (The proof is left as an exercise for the readers).
Thus, max 1 ≤ j ≤ 9 9 8 ( j − 1 9 9 8 ) = ( 9 9 8 / 2 9 9 8 ) ⇒ j − 1 = 2 9 9 8 = 4 9 9 ⇒ j = 5 0 0 .
And so max { ∣ a 1 ∣ , ∣ a 2 ∣ , … , ∣ a 9 9 9 ∣ } = ( 5 0 0 − 1 ) ! ( 9 9 9 − 5 0 0 ) ! 1 = ( 4 9 9 ! ) 2 1 . Hence, Q = 4 9 9 .
Footnote : Can you generalize this?