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Calculus Level 5

Given that ∑ a = 0 ∞ ∑ b = 0 ∞ ∑ c = 0 ∞ a b b c a ! × b ! × c ! = A \displaystyle \large \sum _{a=0}^{\infty }\sum _{b=0}^{ \infty}\sum _{c=0}^{\infty }\frac{a^bb^c}{a!\times b!\times c!}=A Find ⌊ A ⌋ \left\lfloor A \right\rfloor .

For the purpose of this question, assume 0 0 = 1 0^0=1

Inspiration .


The answer is 3814279.

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Julian Poon by Julian Poon , 20, Singapore

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1 solution

Ariel Gershon
May 24, 2016

∑ a = 0 ∞ ∑ b = 0 ∞ ∑ c = 0 ∞ a b b c a ! b ! c ! \sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \dfrac{a^b b^c}{a! b! c!} = ∑ a = 0 ∞ ∑ b = 0 ∞ a b a ! b ! ∑ c = 0 ∞ b c c ! = \sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \dfrac{a^b}{a! b!} \sum_{c=0}^{\infty} \dfrac{b^c}{c!} = ∑ a = 0 ∞ ∑ b = 0 ∞ a b a ! b ! e b = \sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \dfrac{a^b}{a! b!} e^b = ∑ a = 0 ∞ 1 a ! ∑ b = 0 ∞ ( a e ) b b ! = \sum_{a=0}^{\infty} \dfrac{1}{a!} \sum_{b=0}^{\infty} \dfrac{(ae)^b}{b!} = ∑ a = 0 ∞ 1 a ! e e a = \sum_{a=0}^{\infty} \dfrac{1}{a!} e^{ea} = ∑ a = 0 ∞ ( e e ) a a ! = \sum_{a=0}^{\infty} \dfrac{(e^e)^a}{a!} = e e e = \large{e^{e^e}}

Therefore ⌊ A ⌋ = 3814279 \lfloor A \rfloor = 3814279

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Moderator note:

Great approach! It can be helpful to isolate one variable and keep the rest fixed, to find the summation in each term.

Exactly the same.

Aakash Khandelwal - 5 years ago

A A = 3,814,279.104760154

Aakash Khandelwal - 5 years ago

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