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# Fun for me & Challange for you

Let's define an recurrence relation , $${ a }_{ n+1 }=\left( \cfrac { n+2 }{ n } \right) { a }_{ n }\quad \forall n\in N$$ . Such that $${ a }_{ 1 }=2$$ .

Then Find closed form of, $$\displaystyle{f^{ k }\left( n \right) =\underbrace { \sum _{ n=1 }^{ n }{ } \sum _{ n=1 }^{ n }{ } ..........\sum _{ n=1 }^{ n }{ \left( { a }_{ n } \right) } }_{ k\quad times } }$$

Nishu

Note by Nishu Sharma
2 years, 5 months ago

## Comments

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Firstly this is not at all a competitive site, it's a site for learning(don't feel bad I am just correcting you).Take this light heartedly, I am not being harsh at all.

Also be very careful in writing multiple summations, observe that you are using n everywhere, that doesn't properly the meaning of the problem please change the variable of the limits.

- 2 years, 5 months ago

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sorrry I'am totally new to this site , and don't know much about it .. okay thanks for clerifying .

And I'am totally new to latex , But still I think there is no need to change variables , It's well understood from my side . Since after solving first summition we got answer in terms of n , and then we restart with n=1 ,2,3 , and again finally get in terms of n . So I don't think it is incorrect... @Ronak Agarwal

- 2 years, 5 months ago

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The answer is:

$\Large 2\times \left( \begin{matrix} n+k+1 \\ k+2 \end{matrix} \right)$

where:

$$\left( \begin{matrix} n \\ r \end{matrix} \right)$$ is n choose r

- 2 years, 5 months ago

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Sorry but why question is incorrect ? I don't think that question is incorrect , may be possible that it is difficult to understand what I try to post , Since I'am new to latex and this is not bookish problem language , It is original question by me. Also Instead i think your answer is incorrect . The closed form I getting is ..

$\displaystyle{2\left( \begin{matrix} n+k \\ k+1 \end{matrix} \right) }$

May be possible that my answer is wrong , Since It is hypothetical question by me , But still i think you should check your's once again ?

- 2 years, 5 months ago

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Sorry, I made a mistake last time. You are right, the limits of summation are correct. But the meaning of $$k$$ is not well defined. What if $$k=0$$? According to me, when $$k=0$$, no summation is done, and hence the $$n^{th}$$ term should be $$a(n)$$.

- 2 years, 5 months ago

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@Raghav Vaidyanathan @Kartik Sharma @Ronak Agarwal etc etc.... It is challange for you guys ! I have recently Joined brilliant and I found Ronak and Raghav as tough competitor .. So it is for you ! Hope you enjoy ... :)

- 2 years, 5 months ago

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