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

- schrodinger

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- Loser66

- freckles

\[a_1=1 \\ a_2=\frac{3}{1}a_1=3(1)=3 \\ a_3=\frac{4}{2}a_2=2a_2=2(3)=6 \\ a_4=\frac{5}{3}a_3=\frac{5}{3}(6)=5(2)=10 \\ 1,3,6,10,... \text{ subtracting term from its previous term gives: } \\ 2,3,4,.. \\ \text{ doing that again we have } 1,1,1,... \\ \text{ so we have something \in the form } a_n=An^2+Bn+C \\ \text{ we could defined } a_0=0 \text{ since \it fits our pattern thingy } \\ \text{ so we could say } C=0 \\ a_n=An^2+Bn\]
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

same thing but in reverse \[a_{n+1}~=~\dfrac{n+2}{n}\frac{n+1}{n-1}\cdots+\frac{1+2}{1} = \frac{(n+2)!}{n!2!}a_1~ =~ \binom{n+2}{2}a_1\]

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## More answers

- freckles

\the thingy you had before was correct since a1 is 1

- freckles

but this is correct to :p

- freckles

I didn't think to do all of that
that is pretty neat

- freckles

oh 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

so you can skip finding A and B the icky way

- ganeshie8

OMG! how triangular numbers popped up all off sudden!

- freckles

I found the first few terms above

- ganeshie8

Ahh i wasn't thinking.. \(\large \binom{n}{2}\) is a triangular number...

- freckles

Show \[\left(\begin{matrix}n+2 \\ 2\end{matrix}\right) \text{ is a triangular number for } n \ge 0\]
\[\frac{(n+2)!}{2!(n+2-2)!}=\frac{(n+2)!}{2!n!}=\frac{(n+2)(n+1)n!}{2n!}=\frac{(n+2)(n+1)}{2}\]

- Loser66

I am sorry. My computer is crazy!!

- Loser66

I 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

just trying to see if the sequence in question can be converted easily to the familiar triangular numbers sequence form \(\large a_{n} = a_{n-1}+n\)

- xapproachesinfinity

i got where ganesh got that formula
\[a_{30}=\frac{31!}{2!29!}\] seems to me a bit of?

- xapproachesinfinity

off*

- xapproachesinfinity

Am i doing some kind of a mistake?

- freckles

just need simplifying \[a_{30}=\frac{31 \cdot 30 \cdot 29!}{2 \cdot 29!}\]

- xapproachesinfinity

yeah i did but does not look like it is giving the right answer

- freckles

31*15

- xapproachesinfinity

not on the choices

- freckles

oh see what you are saying

- xapproachesinfinity

i got the same thing

- xapproachesinfinity

everything seems to me perfectly good given the pattern

- freckles

oh loser made typeo

- freckles

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0CCUQFjAB&url=https%3A%2F%2Fwww.ets.org%2Fs%2Fgre%2Fpdf%2Fpractice_book_math.pdf&ei=a-R1Vbv1N8fusAWXq4LQCg&usg=AFQjCNHO5Z7_9WxufkCzxQj2ggUgjYjnxg&sig2=NHd5ZQO0tokDsYeYHFKMAA&bvm=bv.95039771,d.b2w
question 25

- freckles

choice A is suppose to read (15)(31)

- xapproachesinfinity

oh ok

- freckles

yeah she is studying for the gre
which she told me she was going to ace

- xapproachesinfinity

i'm thinking or your question
showing that (n+1)(n+2)/2 is triangular for n>0

- freckles

(n+1)(n+2)/2 is triangular I thought

- freckles

consecutive integers divided by 2
that is triangular right?

- xapproachesinfinity

(n+1)(n+2)/2=(n+1)+n+......+1

- xapproachesinfinity

yeah seems to be it is

- freckles

ok 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

we 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

\[S(n)=\sum_{i=1}^{n}i \\ S(n+1)=\sum_{i=1}^{n+1}i \\ S(n+1)-S(n)=(n+1)\]
I think it starts something like this

- freckles

hmmm...

- freckles

could look it up but don't want to cheat yet

- xapproachesinfinity

hmm yeah i remember this

- freckles

lol I don't remember

- freckles

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

- xapproachesinfinity

the algebra proof is the gauss's way :)

- freckles

the proving the equality thing was easy by induction
i just have a hard time remembering how to derive formulas like that

- freckles

it does involve a bag of tricks

- xapproachesinfinity

wow that GRE is not that easy haha

- xapproachesinfinity

has a lot of questions that require some deep thinking lol

- freckles

yep
I don't remember it being that hard

- xapproachesinfinity

you took one before?

- freckles

there are questions on here that I'm not sure I can answer

- freckles

yea about a decade ago

- xapproachesinfinity

i see! I have never taking any standardized test yet

- freckles

wow lucky

- xapproachesinfinity

most of what is in that GRE is hard to me lol

- xapproachesinfinity

what is the duration for such test?
clearly this is not the exam it self?

- xapproachesinfinity

it says 170 mins
nearly 2 hour! for all that! that is a really frustrating hehe

- freckles

Yeah. Very frustrating I bet.

- anonymous

Just 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 \(b_n=\dfrac{1}{(n-1)!}c_n\).
\[b_n=\frac{1}{(n-1)!}c_n~~\implies~~b_{n+1}=\frac{1}{n!}c_{n+1}\]
We have
\[\begin{align*}
b_{n+1}&=\frac{1}{n}b_n\\
\frac{1}{n!}c_{n+1}&=\frac{1}{n}\times \frac{1}{(n-1)!}c_n\\
c_{n+1}&=c_n\\
c_{n+1}-c_n&=0
\end{align*}\]
If we were to sum over \(n=1\) to \(n=k-1\), we'd have
\[\sum_{n=1}^{k-1}(c_{n+1}-c_n)=0\]
which is telescoping. The sum reduces to \(c_k-c_1=0\), or \(c_k=c_1\). This will be our closed form.
We have that \(a_n=\dfrac{(n+1)!}{(n-1)!}c_n=n(n+1)c_n\). 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

Another 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
\[F'(x)=\sum_{n\ge1}a_nx^{n-1}~~\implies~~xF'(x)=\sum_{n\ge1}a_nx^n\]
Now,
\[\begin{align*}
a_{n+1}&=a_n+\frac{2}{n}a_n\\\\
\sum_{n\ge1}a_{n+1}x^n&=\sum_{n\ge1}a_nx^n+2\sum_{n\ge1}\frac{a_n}{n}x^n\\\\
\sum_{n\ge1}a_{n+1}x^{n+1}&=x\sum_{n\ge1}a_nx^n+2x\sum_{n\ge1}\frac{a_n}{n}x^n\\\\
a_1x+\sum_{n\ge1}a_{n+1}x^{n+1}&=x\sum_{n\ge1}a_nx^n+2x\sum_{n\ge1}\frac{a_n}{n}x^n+a_1x\\\\
\sum_{n\ge0}a_{n+1}x^{n+1}&=x^2F'(x)+2xF(x)+x\\\\
\sum_{n\ge1}a_nx^n&=x^2F'(x)+2xF(x)+x\\\\
xF'(x)&=x^2F'(x)+2xF(x)+x\\\\
(1-x)F'(x)-2F(x)&=1\\\\
F'(x)-\frac{2}{1-x}F(x)&=\frac{1}{1-x}\end{align*}\]
We have
\[F(x)=\exp\left(-2\int\frac{dx}{1-x}\right)=\exp(2\ln|1-x|)=(1-x)^2\]
\[\begin{align*}
(1-x)^2F'(x)-2(1-x)F(x)&=1-x\\\\
\frac{d}{dx}\left[(1-x)^2F(x)\right]&=1-x\\\\
(1-x)^2F(x)&=x-\frac{1}{2}x^2+C&C=0\text{ since }F(0)=0\\\\
F(x)&=\frac{2x-x^2}{(1-x)^2}\end{align*}\]

- anonymous

Oops, the generating function should be \(F(x)=\dfrac{2x-x^2}{\color{red}2(1-x)^2}\).

- Loser66

Wowwwwwwwwww. I need study harder!!!

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