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

#Calculus #Challenge #Original #Fun-with-Maths

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Ronak Agarwal - 6 years, 1 month ago

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

Nishu sharma - 6 years, 1 month ago

@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 - 6 years, 1 month ago

@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 - 6 years, 1 month ago

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 ?

Nishu sharma - 6 years, 1 month 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 $a(n)$ .

Raghav Vaidyanathan - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$