It's All Factorials

Calculus Level 5

If ∑ J = 2 1001 ( ( 2001 2 ) ! + J ) = a + b ! π 2 b c ! \displaystyle\sum_{J = 2}^{1001} \left(\left(\frac{2001}{2}\right)! + J\right) = a + \dfrac{b!\sqrt{\pi}}{2^b c!} , where gcd ⁡ ( a , b , c ) = 1 \gcd (a,b,c) = 1 , find a + b + c . a + b + c.


The answer is 504500.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
Aug 19, 2018

Using the gamma function Γ ( p ) = ∫ 0 ∞ t p − 1 e − t d t \Gamma(p) = \int_{0}^{\infty} t^{p - 1} e^{-t} dt we obtain:

Γ ( 1 2 ) = ∫ 0 ∞ t − 1 2 e − t d t \Gamma(\dfrac{1}{2}) = \displaystyle\int_{0}^{\infty} t^{-\frac{1}{2}} e^{-t} dt

Let s 2 = t ⟹ 2 s d s = d t ⟹ s^2 = t \implies 2s ds = dt \implies

Γ ( 1 2 ) = 2 ∫ 0 ∞ e − s 2 d s \Gamma(\dfrac{1}{2}) = 2\displaystyle\int_{0}^{\infty} e^{-s^2} ds

Since s is a dummy variable we can write:

( Γ ( 1 2 ) ) 2 = (\Gamma(\dfrac{1}{2}))^2 = 4 ∫ 0 ∞ e − x 2 d x ∫ 0 ∞ e − y 2 d y = 4\displaystyle\int_{0}^{\infty} e^{-x^2} dx \int_{0}^{\infty} e^{-y^2} dy = 4 ∫ 0 ∞ ∫ 0 ∞ e − ( x 2 + y 2 ) d x d y 4\displaystyle\int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2)} dx dy

Let x = r cos ⁡ θ , y = r sin ⁡ θ x = r\cos\theta, y = r\sin\theta

Using the Jacobian ⟹ ( Γ ( 1 2 ) ) 2 = 4 ∫ 0 π 2 ∫ 0 ∞ e − r 2 r d r d θ = \implies (\Gamma(\dfrac{1}{2}))^2 = 4\displaystyle\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-r^2} r dr d\theta = − 2 ∫ 0 π 2 e − r 2 ∣ 0 ∞ d θ = 2 ∫ 0 π 2 d θ = π ⟹ Γ ( 1 2 ) = π -2\displaystyle\int_{0}^{\frac{\pi}{2}} e^{-r^2}|_{0}^{\infty} d\theta = 2\displaystyle\int_{0}^{\frac{\pi}{2}} d\theta = \pi \implies \Gamma(\dfrac{1}{2}) = \sqrt{\pi}

Now, using integration by parts you can show that Γ ( p + 1 ) = p Γ ( p ) \Gamma(p + 1) = p \Gamma(p)

Show Γ ( 1 ) = 1 \Gamma(1) = 1 and recursively using Γ ( p + 1 ) = p Γ ( p ) ⟹ Γ ( p + 1 ) = p ! \Gamma(p + 1) = p \Gamma(p) \implies \Gamma(p + 1) = p! for any integer p ≥ 0 p \geq 0 and p can be extended to reals so that Γ ( 1 2 ) = Γ ( − 1 2 + 1 ) = π = ( − 1 2 ) ! \Gamma(\dfrac{1}{2}) = \Gamma(\dfrac{-1}{2} + 1) = \sqrt{\pi} = (\dfrac{-1}{2})!

Using Γ ( p + 1 ) = p Γ ( p ) \Gamma(p + 1) = p \Gamma(p) and Γ ( 1 2 ) = π = ( − 1 2 ) ! \Gamma(\dfrac{1}{2}) = \sqrt{\pi} = (\dfrac{-1}{2})!

⟹ \implies

S 0 = Γ ( 3 2 ) = 1 2 ∗ Γ ( 1 2 ) = π 2 = ( 1 2 ) ! S_{0} = \Gamma(\dfrac{3}{2}) = \dfrac{1}{2} * \Gamma(\dfrac{1}{2}) = \dfrac{\sqrt{\pi}}{2} = (\dfrac{1}{2})!

S 1 = Γ ( 5 2 ) = 3 2 ∗ Γ ( 3 2 ) = 1 ∗ 3 ∗ π 2 2 = ( 3 2 ) ! S_{1} = \Gamma(\dfrac{5}{2}) = \dfrac{3}{2} * \Gamma(\dfrac{3}{2}) = \dfrac{1 * 3 * \sqrt{\pi}}{2^2} = (\dfrac{3}{2})!

S 2 = Γ ( 7 2 ) = 5 2 ∗ Γ ( 5 2 ) = 1 ∗ 3 ∗ 5 ∗ π 2 3 = ( 5 2 ) ! S_{2} = \Gamma(\dfrac{7}{2}) = \dfrac{5}{2} * \Gamma(\dfrac{5}{2}) = \dfrac{1 * 3 * 5 * \sqrt{\pi}}{2^3} = (\dfrac{5}{2})!

In General:

S K = Γ ( 2 K + 3 2 ) = 1 ∗ 3 ∗ 5 ∗ ∗ ∗ ( 2 K + 1 ) ∗ π 2 K + 1 = ( K + 1 2 ) ! S_{K} = \Gamma(\dfrac{2K + 3}{2}) = \dfrac{1 * 3 * 5 * * * (2K + 1) * \sqrt{\pi}}{2^{K + 1}} = (K + \frac{1}{2})!

⟹ ( K + 1 2 ) ! = ( 2 K + 1 ) ! ∗ π 2 2 K + 1 ∗ K ! \implies (K + \dfrac{1}{2})! = \dfrac{(2K + 1)! * \sqrt{\pi}}{2^{2K + 1} * K!} , where K K is a non-negative integer.

Note: ( 2 K + 1 ) ! = 1 ∗ 2 ∗ 3 ∗ ∗ ∗ ( 2 K ) ∗ ( 2 K + 1 ) = 2 K ∗ K ! ∗ ( 1 ∗ 3 ∗ 5 ∗ ∗ ∗ ( 2 K + 1 ) ) ⟹ (2K + 1)! = 1 * 2 * 3 * * * (2K) * (2K + 1) = 2^K * K! * (1 * 3 * 5 * * * (2K + 1)) \implies 1 ∗ 3 ∗ 5 ∗ ∗ ∗ ( 2 K + 1 ) = ( 2 K + 1 ) ! 2 K ∗ K ! 1 * 3 * 5 * * * (2K + 1) = \dfrac{(2K + 1)!}{2^K * K!}

For K ∈ N K \in \mathbb{N}

∑ J = 2 K + 1 ( ( K + 1 2 ) ! + J ) = ( K + 2 ) ( K + 1 ) 2 − 1 + K ∗ ( K + 1 2 ) ! = K ( K + 3 ) 2 + ( 2 K + 1 ) ! π 2 2 K + 1 ∗ ( K − 1 ) ! \displaystyle\sum_{J = 2}^{K + 1} ( (K + \dfrac{1}{2})! + J ) = \dfrac{(K + 2)(K + 1)}{2} - 1 + K * (K + \frac{1}{2})! = \dfrac{K(K + 3)}{2} + \frac{(2K + 1)! \sqrt{\pi}}{2^{2K + 1} * (K - 1)!}

∴ ∑ 2 1001 ( ( 1000 + 1 2 ) ! + J ) = 500 ∗ 1003 + ( 2001 ) ! ∗ π 2 2001 ∗ ( 999 ) ! \therefore \displaystyle\sum_{2}^{1001} ((1000 + \dfrac{1}{2})! + J) = 500 * 1003 + \dfrac{(2001)! * \sqrt{\pi}}{2^{2001} * (999)!}
= 501500 + ( 2001 ) ! ∗ π 2 2001 ∗ ( 999 ) ! = a + b ! ∗ π 2 b ∗ c ! ⟹ = 501500 + \dfrac{(2001)! * \sqrt{\pi}}{2^{2001} * (999)!} = a + \dfrac{b! * \sqrt{\pi}}{2^b * c!} \implies a + b + c = 504500 . a + b + c = \boxed{504500}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You're forgetting to multiply the last term in that 2nd line from the bottom by 1000, so that c shouild be 1000, not 999. The correct answer should have been 504501.

Michael Mendrin - 2 years, 9 months ago

Log in to reply

I'm sorry, but I don't see it.

In general I arrived at:

∑ J = 2 K + 1 ( ( K + 1 2 ) ! + J ) = ( K + 2 ) ( K + 1 ) 2 − 1 + K ∗ ( K + 1 2 ) ! = K ( K + 3 ) 2 + ( 2 K + 1 ) ! π 2 2 K + 1 ∗ ( K − 1 ) ! \displaystyle\sum_{J = 2}^{K + 1} ( (K + \dfrac{1}{2})! + J ) = \dfrac{(K + 2)(K + 1)}{2} - 1 + K * (K + \frac{1}{2})! = \dfrac{K(K + 3)}{2} + \frac{(2K + 1)! \sqrt{\pi}}{2^{2K + 1} * (K - 1)!} where, ( K + 1 2 ) ! = ( 2 K + 1 ) ! ∗ π 2 2 K + 1 ∗ K ! (K + \dfrac{1}{2})! = \dfrac{(2K + 1)! * \sqrt{\pi}}{2^{2K + 1} * K!} .

I don't see anything wrong here.

Rocco Dalto - 2 years, 9 months ago

Log in to reply

oh never mind I see my own mistake...dumb oversight right at the very end.

Michael Mendrin - 2 years, 9 months ago

Log in to reply

@Michael Mendrin – No problem.

Rocco Dalto - 2 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...