The Two who Reccur

Probability Level 4

Let there be two sequences defined as

{ a n a_n } = ∑ j = 1 n ∑ k = j n ( n k ) ( k j ) \large \sum_{j=1}^n \sum_{k=j}^n \dbinom n k \dbinom k j

and { b n b_n } = { a n a_n } + 2 n + 1 2^{n+1}

Then { b n + 1 b_{n+1} } can be given by the following recurrence formula

{ b n + 1 b_{n+1} } = α \alpha { b n b_n } + β \beta { b n − 1 b_{n-1} } where α \alpha and β \beta are constant integers. Find α \alpha + β \beta

Hint: { a n + 1 a_{n+1} } can also be given by the same recurrence formula


The answer is -1.

Somyaneel Sinha by Somyaneel Sinha , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Aryan Goyat
Feb 25, 2016

the above summation when we put n=1 can be seen as selecting 1 unique object from n objects lets mark it as 'A' and any number of objects from n-1 objects =2^(n-1) but 'A' can be any out of n objects so it becomes (nC1)2^(n-1) so for n=2 it become (nC2)2^(n-2) and so on ... it become expansion of (x+1)^(n) with 1st term missing and x=2

=3^(n)-2^(n)

an=3^(n)-2^(n)

bn=3^n+2^n

so we get

Alpha=5 and Beta=-6

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Great work friend! Liked the way you used combinatorial argument to calculate the summation. Thanks for the solution.

Somyaneel Sinha - 5 years, 3 months ago

Log in to reply

if you really liked then upvote!

aryan goyat - 5 years, 3 months ago

Log in to reply

Already upvoted your solution!

Somyaneel Sinha - 5 years, 3 months ago
Archit Agrawal
Feb 24, 2016

an=3^n-2^n and bn=3^n+2^n. Alpha=5 and Beta=-6.

0 Helpful 0 Interesting 1 Brilliant 0 Confused

https://brilliant.org/discussions/thread/brilliant-mechanics-contest-season-1/?sort=new check this out.

aryan goyat - 5 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...