A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
Let \(\{a_n\}_{n=1}^\infty \) be defined recursively by \(a_1=1\) and \(a_{n+1}=\dfrac{n+2}{n}a_n\) for \(n\geq 1\)/ Then \(a_{30}\) is equal to??
A) (15)(30)
B) (30)(31)
C) \(\dfrac{31}{29}\)
D) \(\dfrac{32}{30}\)
E) \(\dfrac{32!}{30!2!}\)
Please, help
Loser66
 one year ago
Let \(\{a_n\}_{n=1}^\infty \) be defined recursively by \(a_1=1\) and \(a_{n+1}=\dfrac{n+2}{n}a_n\) for \(n\geq 1\)/ Then \(a_{30}\) is equal to?? A) (15)(30) B) (30)(31) C) \(\dfrac{31}{29}\) D) \(\dfrac{32}{30}\) E) \(\dfrac{32!}{30!2!}\) Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui @Michele_Laino

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[a_1=1 \\ a_2=\frac{3}{1}a_1=3(1)=3 \\ a_3=\frac{4}{2}a_2=2a_2=2(3)=6 \\ a_4=\frac{5}{3}a_3=\frac{5}{3}(6)=5(2)=10 \\ 1,3,6,10,... \text{ subtracting term from its previous term gives: } \\ 2,3,4,.. \\ \text{ doing that again we have } 1,1,1,... \\ \text{ so we have something \in the form } a_n=An^2+Bn+C \\ \text{ we could defined } a_0=0 \text{ since \it fits our pattern thingy } \\ \text{ so we could say } C=0 \\ a_n=An^2+Bn\] you can use the other terms in the sequence to find A and B this is the way I would think to do it there may be a quicker way (don't know about that though)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1same thing but in reverse \[a_{n+1}~=~\dfrac{n+2}{n}\frac{n+1}{n1}\cdots+\frac{1+2}{1} = \frac{(n+2)!}{n!2!}a_1~ =~ \binom{n+2}{2}a_1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\the thingy you had before was correct since a1 is 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.5but this is correct to :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.5I didn't think to do all of that that is pretty neat

freckles
 one year ago
Best ResponseYou've already chosen the best response.5oh oh if you notice that the first term is 1 the second terms is 1+2 the third term is 1+2+3 the fourth term is 1+2+3+4 ... then you know the nth term is given by n(n+1)/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.5so you can skip finding A and B the icky way

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1OMG! how triangular numbers popped up all off sudden!

freckles
 one year ago
Best ResponseYou've already chosen the best response.5I found the first few terms above

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh i wasn't thinking.. \(\large \binom{n}{2}\) is a triangular number...

freckles
 one year ago
Best ResponseYou've already chosen the best response.5Show \[\left(\begin{matrix}n+2 \\ 2\end{matrix}\right) \text{ is a triangular number for } n \ge 0\] \[\frac{(n+2)!}{2!(n+22)!}=\frac{(n+2)!}{2!n!}=\frac{(n+2)(n+1)n!}{2n!}=\frac{(n+2)(n+1)}{2}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I am sorry. My computer is crazy!!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have a formula to find it, but I don't have my note with me right now. I will post it when I get home

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1just trying to see if the sequence in question can be converted easily to the familiar triangular numbers sequence form \(\large a_{n} = a_{n1}+n\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i got where ganesh got that formula \[a_{30}=\frac{31!}{2!29!}\] seems to me a bit of?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0Am i doing some kind of a mistake?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5just need simplifying \[a_{30}=\frac{31 \cdot 30 \cdot 29!}{2 \cdot 29!}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0yeah i did but does not look like it is giving the right answer

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0not on the choices

freckles
 one year ago
Best ResponseYou've already chosen the best response.5oh see what you are saying

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i got the same thing

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0everything seems to me perfectly good given the pattern

freckles
 one year ago
Best ResponseYou've already chosen the best response.5choice A is suppose to read (15)(31)

freckles
 one year ago
Best ResponseYou've already chosen the best response.5yeah she is studying for the gre which she told me she was going to ace

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i'm thinking or your question showing that (n+1)(n+2)/2 is triangular for n>0

freckles
 one year ago
Best ResponseYou've already chosen the best response.5(n+1)(n+2)/2 is triangular I thought

freckles
 one year ago
Best ResponseYou've already chosen the best response.5consecutive integers divided by 2 that is triangular right?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0(n+1)(n+2)/2=(n+1)+n+......+1

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0yeah seems to be it is

freckles
 one year ago
Best ResponseYou've already chosen the best response.5ok you want to prove: \[\sum_{i=1}^{n}i=\frac{n(n+1)}{2} \\ \sum_{i=1}^{n+1}i=\frac{(n+1)(n+1+1)}{2}=\frac{(n+1)(n+2)}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5we can prove that by induction there is also one other way and that is to actually derive that equality sometimes I forget how to do that let me see I can derive it... *freckle's processor processing*

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[S(n)=\sum_{i=1}^{n}i \\ S(n+1)=\sum_{i=1}^{n+1}i \\ S(n+1)S(n)=(n+1)\] I think it starts something like this

freckles
 one year ago
Best ResponseYou've already chosen the best response.5could look it up but don't want to cheat yet

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm yeah i remember this

freckles
 one year ago
Best ResponseYou've already chosen the best response.5http://www.maths.surrey.ac.uk/hostedsites/R.Knott/runsums/triNbProof.html

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0the algebra proof is the gauss's way :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.5the proving the equality thing was easy by induction i just have a hard time remembering how to derive formulas like that

freckles
 one year ago
Best ResponseYou've already chosen the best response.5it does involve a bag of tricks

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0wow that GRE is not that easy haha

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0has a lot of questions that require some deep thinking lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.5yep I don't remember it being that hard

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0you took one before?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5there are questions on here that I'm not sure I can answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.5yea about a decade ago

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i see! I have never taking any standardized test yet

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0most of what is in that GRE is hard to me lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0what is the duration for such test? clearly this is not the exam it self?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0it says 170 mins nearly 2 hour! for all that! that is a really frustrating hehe

freckles
 one year ago
Best ResponseYou've already chosen the best response.5Yeah. Very frustrating I bet.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just offering another approach we can take. First set \(a_n=\dfrac(n+1)!b_n\). Then \[a_n=(n+1)!b_n~~\implies~~a_{n+1}=(n+2)!\,b_{n+1}\] So, \[\begin{align*} a_{n+1}&=\frac{n+2}{n}a_n\\ (n+2)!\,b_{n+1}&=\frac{(n+2)!}{n}b_n\\ b_{n+1}&=\frac{1}{n}b_n \end{align*}\] Next set \(b_n=\dfrac{1}{(n1)!}c_n\). \[b_n=\frac{1}{(n1)!}c_n~~\implies~~b_{n+1}=\frac{1}{n!}c_{n+1}\] We have \[\begin{align*} b_{n+1}&=\frac{1}{n}b_n\\ \frac{1}{n!}c_{n+1}&=\frac{1}{n}\times \frac{1}{(n1)!}c_n\\ c_{n+1}&=c_n\\ c_{n+1}c_n&=0 \end{align*}\] If we were to sum over \(n=1\) to \(n=k1\), we'd have \[\sum_{n=1}^{k1}(c_{n+1}c_n)=0\] which is telescoping. The sum reduces to \(c_kc_1=0\), or \(c_k=c_1\). This will be our closed form. We have that \(a_n=\dfrac{(n+1)!}{(n1)!}c_n=n(n+1)c_n\). Given that \(a_1=1\), we find \[1=\dfrac{2!}{0!}c_1~~\implies~~c_1=\frac{1}{2}\] which gives the closed form \[a_n=\frac{n(n+1)}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another approach, just because :) \[a_{n+1}=\frac{n+2}{n}a_n=a_n+\frac{2}{n}a_n\] Let \(F(x)=\sum\limits_{n\ge1}\dfrac{a_n}{n}x^n\) denote the generating function for \(\dfrac{1}{n}a_n\). We have \[F'(x)=\sum_{n\ge1}a_nx^{n1}~~\implies~~xF'(x)=\sum_{n\ge1}a_nx^n\] Now, \[\begin{align*} a_{n+1}&=a_n+\frac{2}{n}a_n\\\\ \sum_{n\ge1}a_{n+1}x^n&=\sum_{n\ge1}a_nx^n+2\sum_{n\ge1}\frac{a_n}{n}x^n\\\\ \sum_{n\ge1}a_{n+1}x^{n+1}&=x\sum_{n\ge1}a_nx^n+2x\sum_{n\ge1}\frac{a_n}{n}x^n\\\\ a_1x+\sum_{n\ge1}a_{n+1}x^{n+1}&=x\sum_{n\ge1}a_nx^n+2x\sum_{n\ge1}\frac{a_n}{n}x^n+a_1x\\\\ \sum_{n\ge0}a_{n+1}x^{n+1}&=x^2F'(x)+2xF(x)+x\\\\ \sum_{n\ge1}a_nx^n&=x^2F'(x)+2xF(x)+x\\\\ xF'(x)&=x^2F'(x)+2xF(x)+x\\\\ (1x)F'(x)2F(x)&=1\\\\ F'(x)\frac{2}{1x}F(x)&=\frac{1}{1x}\end{align*}\] We have \[F(x)=\exp\left(2\int\frac{dx}{1x}\right)=\exp(2\ln1x)=(1x)^2\] \[\begin{align*} (1x)^2F'(x)2(1x)F(x)&=1x\\\\ \frac{d}{dx}\left[(1x)^2F(x)\right]&=1x\\\\ (1x)^2F(x)&=x\frac{1}{2}x^2+C&C=0\text{ since }F(0)=0\\\\ F(x)&=\frac{2xx^2}{(1x)^2}\end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops, the generating function should be \(F(x)=\dfrac{2xx^2}{\color{red}2(1x)^2}\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Wowwwwwwwwww. I need study harder!!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.