Staggered triplets

Calculus Level 4

S = 1 1 ⋅ 3 ⋅ 5 + 1 2 ⋅ 4 ⋅ 6 + 1 3 ⋅ 5 ⋅ 7 + 1 4 ⋅ 6 ⋅ 8 + 1 5 ⋅ 7 ⋅ 9 + ⋯ S=\dfrac { 1 }{ 1\cdot 3\cdot 5 } +\dfrac { 1 }{ 2\cdot 4\cdot 6 } +\dfrac { 1 }{ 3\cdot 5\cdot 7 } +\dfrac { 1 }{ 4\cdot 6\cdot 8 } +\dfrac { 1 }{ 5\cdot 7\cdot 9 } + \cdots

If S S can be expressed in the form A B C D \frac { A }{ { B }^{ C }D } with A A , B B , C C , and D D being positive prime integers, find the value of A + B + C + D D . \dfrac { A+B+C+D }{ D }.

This problem is original.


The answer is 7.

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Jonas Katona by Jonas Katona , 22, USA

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3 solutions

Manuel Kahayon
Dec 15, 2015

Seriously??? I don't even know what calculus is and yet I answered it... Anyways...

all of the above fractions can be expressed in the form 1 ( n − 2 ) ( n ) ( n + 2 ) \frac {1} {(n-2)(n)(n+2)} , or 1 4 ∗ ( 1 ( n − 2 ) ( n ) − 1 ( n ) ( n + 2 ) ) \frac {1} {4} * (\frac {1} {(n-2)(n)} - \frac {1} {(n)(n+2)})

This simplifies to 1 4 ∗ ( 1 ( 1 ) ( 3 ) − 1 ( 3 ) ( 5 ) + 1 ( 2 ) ( 4 ) − 1 ( 4 ) ( 6 ) + 1 ( 3 ) ( 5 ) − 1 ( 5 ) ( 7 ) + 1 ( 4 ) ( 6 ) − 1 ( 6 ) ( 8 ) … ) \frac {1} {4} * (\frac {1} {(1)(3)} - \frac {1} {(3)(5)} + \frac {1} {(2)(4)} - \frac {1} {(4)(6)} + \frac {1} {(3)(5)} - \frac {1} {(5)(7)} + \frac {1} {(4)(6)} - \frac {1} {(6)(8)} \ldots)

Which telescopes to 1 4 ∗ ( 1 ( 1 ) ( 3 ) + 1 ( 2 ) ( 4 ) ) \frac{1} {4} * (\frac {1} {(1)(3)} + \frac {1} {(2)(4)})

This equals to 11 96 \frac {11} {96} or 11 2 5 ∗ 3 \frac {11} {2^5*3}

Therefore, A = 11 , B = 2 , C = 5 , D = 3 A=11, B=2, C=5, D=3

Then, the required answer is 11 + 2 + 5 + 3 3 = 7 \frac {11 + 2 + 5 +3} {3} = 7

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Your solution was very clean. Nice job! Because of your solution, I have changed the category frim calculus to algebra. This is a telescoping series.

Jonas Katona - 5 years, 6 months ago

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I enjoyed that problem a lot Jonas! Was a little messy but had a nice conclusion.

Adrian Castro - 5 years, 6 months ago

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Thank you very much, Adrian! And I know, most of my problems are pretty messy. But that's what I try to do: the problems look simple, but are actually pretty difficult. Either way, the final result is usually pretty elegant.

Jonas Katona - 5 years, 6 months ago

I did the same. How could this be under calculus earlier?

Shreyash Rai - 5 years, 5 months ago

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I have no idea, but someone (probably an admin) changed the category back to calculus.

I guess partial fractions are now considered calculus? But then why did they change my "a growth of triplets" to algebra? It's a debatable subject, I guess...But fundamentally, at least these two of my series questions require no calculus knowledge.

I'm changing it back to algebra. Lol ~

Jonas Katona - 5 years, 5 months ago

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Wait, no, jk, because they put it under a calculus wiki page...

Jonas Katona - 5 years, 5 months ago

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@Jonas Katona – Weird. Now both are under calculus. Well nice problem anyway

Shreyash Rai - 5 years, 5 months ago

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@Shreyash Rai – Hahaha thank you!

Jonas Katona - 5 years, 5 months ago

Same solution as @Manuel Kahayon .

Vignesh S - 5 years, 2 months ago
Chew-Seong Cheong
Dec 16, 2015

S = ∑ n = 1 ∞ 1 n ( n + 2 ) ( n + 4 ) = 1 8 ∑ n = 1 ∞ ( 1 n − 2 n + 2 + 1 n + 4 ) = 1 8 ∑ n = 1 ∞ ( 1 n − 1 n + 2 − 1 n + 2 + 1 n + 4 ) = 1 8 ( ∑ n = 1 ∞ 1 n − ∑ n = 3 ∞ 1 n − ∑ n = 3 ∞ 1 n + ∑ n = 5 ∞ 1 n ) = 1 8 ( 1 1 + 1 2 − 1 3 − 1 4 ) = 1 8 ( 12 + 6 − 4 − 3 12 ) = 11 96 = 11 2 5 3 \begin{aligned} S & = \sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} \\ & = \frac{1}{8} \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{2}{n+2} + \frac{1}{n+4} \right) \\ & = \frac{1}{8} \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+2} - \frac{1}{n+2} + \frac{1}{n+4} \right) \\ & = \frac{1}{8} \left( \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=3}^\infty \frac{1}{n} - \sum_{n=3}^\infty \frac{1}{n} + \sum_{n=5}^\infty \frac{1}{n} \right) \\ & = \frac{1}{8} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} \right) = \frac{1}{8} \left(\frac{12+6-4-3}{12}\right) = \frac{11}{96} = \frac{11}{2^53} \end{aligned}

⇒ A + B + C + D D = 11 + 2 + 5 + 3 3 = 21 3 = 7 \Rightarrow \dfrac{A+B+C+D}{D} = \dfrac{11+2+5+3}{3} = \dfrac{21}{3} = \boxed{7}

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S = ∑ n = 3 ∞ 1 ( n − 2 ) n ( n + 2 ) = 1 8 ( ∑ n = 3 ∞ 1 n − 2 − 2 ∗ ∑ n = 3 ∞ 1 n + ∑ n = 3 ∞ 1 n + 2 ) = 1 8 ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + . . . − 2 3 − 2 4 − 2 5 − 2 6 − 2 7 − . . . + 1 5 + 1 6 + 1 7 + . . . ) = 1 8 ( 1 1 + 1 2 − 1 3 − 1 4 + 0 + 0... ) = 1 8 ∗ 11 12 = 11 2 5 ∗ 3 = A B C ∗ D ∴ A + B + C + D D = 11 + 2 + 5 + 3 3 = 7 \displaystyle S=\sum_{n=3}^{\infty}\dfrac 1 {(n-2)n(n+2)}=\dfrac1 8 \left ( \sum_{n=3}^{\infty}\dfrac 1 {n - 2} - 2*\sum_{n=3}^{\infty}\dfrac 1 n +\sum_{n=3}^{\infty}\dfrac 1 {n+ 2} \right ) \\ =\dfrac1 8 \left ( \color{#3D99F6}{\dfrac 1 1 +\dfrac 1 2 +\dfrac 1 3 +\dfrac 1 4} +\dfrac 1 5 +\dfrac 1 6 +\dfrac 1 7 +. . . {\Large \color{#D61F06}{ - }}\color{#3D99F6}{ \dfrac 2 3} {\Large \color{#D61F06}{ - }}\color{#3D99F6}{ \dfrac 2 4} {\Large \color{#D61F06}{ - }} \dfrac 2 5{\Large \color{#D61F06}{ - }} \dfrac 2 6 {\Large \color{#D61F06}{ - }} \dfrac 2 7 {\Large \color{#D61F06}{ - }} . . . +\dfrac 1 5 +\dfrac 1 6 +\dfrac 1 7 + . . .\right) \\ =\dfrac1 8 \left ( \dfrac 1 1 +\dfrac 1 2 - \dfrac 1 3 - \dfrac 1 4 +0+0 . . .\right) \\ =\dfrac1 8*\dfrac {11}{12}=\dfrac {11}{2^5*3}=\dfrac A {B^C*D}\\ \therefore ~\dfrac{A+B+C+D}{D}=\dfrac{11+2+5+3}{3}= \huge \color{#EC7300}{7}

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