A calculus problem by A Former Brilliant Member
p = 1 ∑ 1 5 ( p + 2 ) ! p ⋅ 2 p = w − z ! x y
The equation above holds true for some positive integers w , x , y , and z . Evaluate ∣ ∣ ∣ ∣ w y x z ∣ ∣ ∣ ∣ .
You can try more of my problems here .
The answer is -15.
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1 solution
Did the same! BTW, I was thinking what is the closed form of ∑ p = 0 n p ! x p and on comparing it a little, with ∑ p = 0 n P ( n , p ) , I found it to be n ! e x Γ ( n + 1 , x ) which seemed interesting though to me and I solved it with that too. *Just wanted to share that result!
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Wow !! It didn't occur to me to think laterally the way you did . Nice observation skills Kartik :)
It'd be nice if you could post the other approach as a new solution, everyone will benefit from a new unconventional method using Gamma Function .
Please add to the problem that w , x , y , z are all integers. There are infinitely many solutions otherwise
Be careful and specify that x must be squarefree. Otherwise the problem has multiple solutions. For example -47.
You should mention in the question that x is a prime or it is square free otherwise there will be many more solutions Nice question btw!
Why is this under calculus . This is an hardcore algebra problem.
This should be an algebra category. Calculus is of no use here .
( p + 2 ) ! p ⋅ 2 p = ( p + 2 ) ! ( p + 2 − 2 ) ⋅ 2 p = ( p + 1 ) ! 2 p − ( p + 2 ) ! 2 p + 1 .
Therefore our required sum is p = 1 ∑ 1 5 ( p + 2 ) ! p ⋅ 2 p = p = 1 ∑ 1 5 ( p + 1 ) ! 2 p − ( p + 2 ) ! 2 p + 1 , which by the application of Telescoping Series evaluates to 1 − 1 7 ! 2 1 6 .
Hence , w = 1 , x = 2 , y = 1 6 and z = 1 7 .
So we need to evaluate ∣ ∣ ∣ ∣ 1 1 6 2 1 7 ∣ ∣ ∣ ∣ which comes out to be − 1 5 .