Question

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

Answer 1

@Kainui @Michele_Laino

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

but this is correct to :p

Answer 6

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

Answer 7

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

Answer 8

so you can skip finding A and B the icky way

Answer 9

OMG! how triangular numbers popped up all off sudden!

Answer 10

I found the first few terms above

Answer 11

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

Answer 12

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

Answer 13

I am sorry. My computer is crazy!!

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

off*

Answer 18

Am i doing some kind of a mistake?

Answer 19

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

Answer 20

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

Answer 21

31*15

Answer 22

not on the choices

Answer 23

oh see what you are saying

Answer 24

i got the same thing

Answer 25

everything seems to me perfectly good given the pattern

Answer 26

oh loser made typeo

Answer 27

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

Answer 28

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

Answer 29

oh ok

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

consecutive integers divided by 2
that is triangular right?

Answer 34

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

Answer 35

yeah seems to be it is

Answer 36

ok you want to prove:
\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2} \\ \sum_{i=1}^{n+1}i=\frac{(n+1)(n+1+1)}{2}=\frac{(n+1)(n+2)}{2}\]

Answer 37

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*

Answer 38

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

Answer 39

hmmm...

Answer 40

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

Answer 41

hmm yeah i remember this

Answer 42

lol I don't remember

Answer 43

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

Answer 44

the algebra proof is the gauss's way :)

Answer 45

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

Answer 46

it does involve a bag of tricks

Answer 47

wow that GRE is not that easy haha

Answer 48

has a lot of questions that require some deep thinking lol

Answer 49

yep
I don't remember it being that hard

Answer 50

you took one before?

Answer 51

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

Answer 52

yea about a decade ago

Answer 53

i see! I have never taking any standardized test yet

Answer 54

wow lucky

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

Yeah. Very frustrating I bet.

Answer 59

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
\[a_n=\frac{n(n+1)}{2}\]

Answer 60

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
\[F(x)=\exp\left(-2\int\frac{dx}{1-x}\right)=\exp(2\ln|1-x|)=(1-x)^2\]
\[\begin{align*}
(1-x)^2F'(x)-2(1-x)F(x)&=1-x\\\\
\frac{d}{dx}\left[(1-x)^2F(x)\right]&=1-x\\\\
(1-x)^2F(x)&=x-\frac{1}{2}x^2+C&C=0\text{ since }F(0)=0\\\\
F(x)&=\frac{2x-x^2}{(1-x)^2}\end{align*}\]

Answer 61

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

Answer 62

Wowwwwwwwwww. I need study harder!!!