Une Somme Difficile

x = 1 + 1 12 + 1 ( 4 ) 2 ! 1 1 2 2 + 1 ( 4 ) ( 7 ) 3 ! 1 1 2 3 + 1 ( 4 ) ( 7 ) ( 10 ) 4 ! 1 1 2 4 + x = 1+\frac{1}{12}+\frac{1(4)}{2!}\frac{1}{12^2}+\frac{1(4)(7)}{3!}\frac{1}{12^3}+\frac{1(4)(7)(10)}{4!}\frac{1}{12^4}+\cdots

Find the value of 3 x 3 3x^3 .


The answer is 4.

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1 solution

Kishlaya Jaiswal
Dec 27, 2013

We might think of applying the Generalized Binomial Theorem , but it can't be applied directly.

So, need to do some modifications with the series so that we can easily see some binomial pattern.

x = 1 + 1 12 + 1 ( 4 ) 2 ! 1 1 2 2 + 1 ( 4 ) ( 7 ) 3 ! 1 1 2 3 + 1 ( 4 ) ( 7 ) ( 10 ) 4 ! 1 1 2 4 + x=1+\frac{1}{12}+\frac{1(4)}{2!}\frac{1}{12^2}+\frac{1(4)(7)}{3!}\frac{1}{12^3}+\frac{1(4)(7)(10)}{4!}\frac{1}{12^4}+\ldots

We'll break 1 2 x 12^x ( x x is any arbitrary power of 12) into 4 x × 3 x 4^x \times 3^x and take 3 x 3^x in the numerator. Observe carefully -

x = 1 + 1 3 × 1 4 + 1 3 × 4 3 2 ! 1 4 2 + 1 3 × 4 3 × 7 3 3 ! 1 4 3 + x=1+\frac{1}{3}\times \frac{1}{4}+\frac{\frac{1}{3}\times \frac{4}{3}}{2!}\frac{1}{4^2}+\frac{\frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}}{3!}\frac{1}{4^3}+\ldots

x = 1 + 1 3 × 1 4 + 1 3 × ( 1 3 1 ) 2 ! 1 4 2 + 1 3 × ( 1 3 ) × ( 1 3 2 ) 3 ! 1 4 3 + x=1+\frac{-1}{3}\times \frac{-1}{4}+\frac{\frac{-1}{3}\times (\frac{-1}{3}-1)}{2!}\frac{1}{4^2}+\frac{\frac{-1}{3}\times (\frac{-1}{3})\times (\frac{-1}{3}-2)}{3!}\frac{-1}{4^3}+\ldots

Now, this is nothing but expansion of x = ( 1 1 4 ) 1 3 ( 3 4 ) 1 3 ( 4 3 ) 1 3 x = (1-\frac{1}{4})^{\frac{-1}{3}} \Rightarrow (\frac{3}{4})^{\frac{-1}{3}} \Rightarrow (\frac{4}{3})^{\frac{1}{3}}

Hence, x = ( 4 3 ) 1 3 x = (\frac{4}{3})^{\frac{1}{3}} . Therefore, 3 x 3 = 4 3x^3 = \boxed{4}

Now why is the question in french?

Joshua Ong - 7 years, 3 months ago

Sir @Calvin Lin can you please change the wording of the problem to English. It was a beautiful problem and I want everyone else to try it.

I guess, the problem statement would something like - If x = 1 + 1 12 + 1 ( 4 ) 2 ! 1 1 2 2 + 1 ( 4 ) ( 7 ) 3 ! 1 1 2 3 + x = 1 + \frac{1}{12} + \frac{1(4)}{2!} \frac{1}{12^2} + \frac{1(4)(7)}{3!} \frac{1}{12^3} + \ldots Then find the value of 3 x 2 3x^2

Kishlaya Jaiswal - 6 years, 3 months ago

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@Kishlaya Jaiswal

Vous possédez de bonnes compétences en effet d'observation :) Je ne aurais jamais fait le chemin que vous avez fait .

Au fait ,je suppose que la question aurait été mieux si ce était comme suit :

x = 1 + 1 12 + 1 ( 2 ) 2 ! 1 1 2 2 + 1 ( 2 ) ( 4 ) 3 ! 1 1 2 3 + 1 ( 2 ) ( 4 ) ( 7 ) 4 ! 1 1 2 4 + \begin{aligned}x&=1+\frac{1}{12}+\frac{1(2)}{2!}\frac{1}{12^2}+\frac{1(2)(4)}{3!}\frac{1}{12^3}\\&+\frac{1(2)(4)(7)}{4!}\frac{1}{12^4}+\dots\end{aligned}

Que pensez-vous ?

A Former Brilliant Member - 6 years, 3 months ago

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fait amusant : Jetez un oeil à la date de cette solution posté ...

(Google Translate is still so incapable)

Kishlaya Jaiswal - 6 years, 3 months ago

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