Jacobsthal-Geometric Series

Algebra Level 5

J n = { 0 if n = 0 ; 1 if n = 1 ; J n − 1 + 2 J n − 2 if n > 1. \text{J}_{n}=\begin{cases} 0 \quad & \text{if}~n=0; \\ 1 \quad & \text{if} ~ n=1; \\ \text{J}_{n-1}+2\text{J}_{n-2} \quad & \text{if}~n>1.\end{cases}

Jacobsthal Numbers are defined by the recurrence relation as described above. Evaluate: ∑ n = 0 ∞ J n 1 0 n + 1 . \displaystyle \sum_{n=0}^{\infty} \dfrac{\text{J}_{n}}{10^{n+1}} \; .

If the value of the above expression is in the form A − 1 A^{-1} , find A A .

Inspiration


The answer is 88.

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Rohit Udaiwal by Rohit Udaiwal , 20, India

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4 solutions

Ayush Verma
Apr 9, 2016

J 0 = 0 , J 1 = 1 & J n = J n − 1 + 2 J n − 2 ⇒ J n 10 n + 1 = 1 10 . J n − 1 10 n + 2 100 . J n − 2 10 n − 1 ⇒ ∑ n = 2 ∞ J n 10 n + 1 = 1 10 ∑ n = 2 ∞ J n − 1 10 n + 2 100 ∑ n = 2 ∞ J n − 2 10 n − 1 ⇒ A − J 1 10 2 − J 0 10 = 1 10 ( A − J 0 10 ) + 2 100 A ⇒ A − 1 100 − 0 = 1 10 ( A − 0 ) + 2 100 A ⇒ 88 100 A = 1 100 ⇒ A − 1 = 88 { J }_{ 0 }=0{ \quad ,\quad J }_{ 1 }=1\\ \\ \& \quad { J }_{ n }={ J }_{ n-1 }+2{ J }_{ n-2 }\quad \Rightarrow \cfrac { { J }_{ n } }{ { 10 }^{ n+1 } } =\cfrac { 1 }{ 10 } .\cfrac { { J }_{ n-1 } }{ { 10 }^{ n } } +\cfrac { 2 }{ 100 } .\cfrac { { J }_{ n-2 } }{ { 10 }^{ n-1 } } \\ \\ \Rightarrow \sum _{ n=2 }^{ \infty }{ \cfrac { { J }_{ n } }{ { 10 }^{ n+1 } } } =\cfrac { 1 }{ 10 } \sum _{ n=2 }^{ \infty }{ \cfrac { { J }_{ n-1 } }{ { 10 }^{ n } } } +\cfrac { 2 }{ 100 } \sum _{ n=2 }^{ \infty }{ \cfrac { { J }_{ n-2 } }{ { 10 }^{ n-1 } } } \\ \\ \Rightarrow A-\cfrac { { J }_{ 1 } }{ { 10 }^{ 2 } } -\cfrac { { J }_{ 0 } }{ { 10 } } =\cfrac { 1 }{ 10 } \left( A-\cfrac { { J }_{ 0 } }{ { 10 } } \right) +\cfrac { 2 }{ 100 } A\\ \\ \Rightarrow A-\cfrac { 1 }{ 100 } -0=\cfrac { 1 }{ 10 } \left( A-0 \right) +\cfrac { 2 }{ 100 } A\\ \\ \Rightarrow \cfrac { 88 }{ 100 } A=\cfrac { 1 }{ 100 } \\ \\ \Rightarrow { A }^{ -1 }=88

10 Helpful 0 Interesting 0 Brilliant 0 Confused

The best way is to use difference equation . That process is more classic. But well done.

Prithwish Mukherjee - 2 years, 3 months ago
Rohit Udaiwal
Apr 8, 2016

Relevant wiki: Recurrence Relations

S = ∑ n = 0 ∞ J n 1 0 n + 1 = 0 10 + 1 1 0 2 + 1 1 0 3 + 3 1 0 4 + 5 1 0 5 + 11 1 0 6 + 21 1 0 8 + … ⟹ S 10 = 1 1 0 3 + 1 1 0 4 + 3 1 0 5 + 5 1 0 6 + 11 1 0 7 + 21 1 0 8 + … ⟹ S + S 10 = 1 1 0 2 + 2 1 0 3 + 4 1 0 4 + … 11 S = 1 10 + 2 1 0 2 + 4 1 0 3 + … [ Infinite GP ] = 1 10 1 − 2 10 = 1 8 ∴ S = 1 88 \begin{aligned} \mathbf{S}= & \sum_{n=0}^{\infty} \dfrac{\text{J}_{n}}{10^{n+1}} \\ = &\dfrac{0}{10}+ \dfrac{1}{10^2} +\dfrac{1}{10^3}+\dfrac{3}{10^4}+\dfrac{5}{10^5}+\dfrac{11}{10^6} +\dfrac{21}{10^8}+\ldots \\ \implies \dfrac{\mathbf{S}}{10}= & \dfrac{1}{10^3} +\dfrac{1}{10^4}+\dfrac{3}{10^5}+\dfrac{5}{10^6}+\dfrac{11}{10^7} +\dfrac{21}{10^8}+\ldots \\ \implies \mathbf{S}+\dfrac{\mathbf{S}}{10}= & \dfrac{1}{10^2}+\dfrac{2}{10^3}+\dfrac{4}{10^4}+\ldots \\ 11\mathbf{S}= & \dfrac{1}{10}+\dfrac{2}{10^2}+\dfrac{4}{10^3}+\ldots \quad \quad [\text{Infinite GP}] \\ = & \dfrac{\frac{1}{10}}{1-\frac{2}{10}}=\dfrac{1}{8} \\ \therefore \mathbf{S}= & \displaystyle \boxed{\dfrac{1}{88}} \end{aligned}

Bonus :Show that 3 J n + ( − 1 ) n = 2 n 3\text{J}_{n}+(-1)^n=2^n ,where J n \text{J}_n denotes the n th n^{\text{th}} Jacobsthal Number.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

On solving the given recurrence, we get :- J n = 1 3 ( 2 n − ( − 1 ) n ) J_n \,=\, \frac{1}{3} \left(2^{n} \,-\,(-1)^{n}\right) . It can easily be seen that 3 ⋅ J n + ( − 1 ) n = 2 n \color{#D61F06}{3 \cdot J_n \,+\, (-1)^{n} \,=\, 2^{n}} .

Aditya Sky - 5 years, 2 months ago

Nice problem! Thanks for citing my problem as an inspiration :)

Manuel Kahayon - 5 years, 2 months ago

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Your problem was very cool.Congratulations,you solved it :)!

Rohit Udaiwal - 5 years, 2 months ago

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Thanks! Although, sadly, it was a ripoff of a similar problem, but the series was defined as a 1 = a 2 = a 3 = 1 , a n = a n − 1 + a n − 2 + a n − 3 a_1=a_2=a_3=1, a_n = a_{n-1}+ a_{n-2}+a_{n-3} for n ≥ 4 n \geq 4

Manuel Kahayon - 5 years, 2 months ago

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@Manuel Kahayon – Oh,that's painful.Nevertheless, could you link me with the problem?

Rohit Udaiwal - 5 years, 2 months ago

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@Rohit Udaiwal – No links. It was a written problem on our training test paper...

Manuel Kahayon - 5 years, 2 months ago

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@Manuel Kahayon – Ohk.!BTW I like your problems a lot.Please keep posting :)

Rohit Udaiwal - 5 years, 2 months ago

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@Rohit Udaiwal – Thanks! (This is an extension of the comment since 1-word comments are not allowed)

Manuel Kahayon - 5 years, 2 months ago

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@Manuel Kahayon – I really think one-word comments should be allowed here. @Andrew Ellinor

Rohit Udaiwal - 5 years, 2 months ago

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@Rohit Udaiwal – Then take it up with me in the Brilliant Lounge :P

Andrew Ellinor - 5 years, 2 months ago
展豪 張
Apr 10, 2016

Let f ( x ) = J 0 + J 1 x + J 2 x 2 + ⋯ f(x)=J_0+J_1x+J_2x^2+\cdots .
From the recurrence relation we have f ( x ) ( 1 − x − 2 x 2 ) = x f(x)(1-x-2x^2)=x
i.e. f ( x ) = x 1 − x − 2 x 2 f(x)=\dfrac x{1-x-2x^2}
The sum = 1 10 f ( 1 10 ) = 1 88 =\dfrac 1{10}f(\dfrac 1{10})=\dfrac 1{88}
A = 88 A=88



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Shourya Pandey
Apr 9, 2016

The link just gives away the formula J n = 1 3 ( 2 n − ( − 1 ) n ) J_{n} = \frac {1}{3}(2^{n} - (-1)^{n})

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks,I've removed that link.

Rohit Udaiwal - 5 years, 2 months ago

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