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

## Comments

Sort by:

TopNewestFirstly 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. – Ronak Agarwal · 2 years, 2 months ago

Log in to reply

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 – Nishu Sharma · 2 years, 2 months ago

Log in to reply

@Nishu sharma

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 – Raghav Vaidyanathan · 2 years, 2 months ago

Log in to reply

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

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 ? – Nishu Sharma · 2 years, 2 months ago

Log in to reply

– Raghav Vaidyanathan · 2 years, 2 months ago

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 \(a(n)\).Log in to reply

@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 ... :) – Nishu Sharma · 2 years, 2 months ago

Log in to reply