## 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

1. Loser66

@Kainui @Michele_Laino

2. 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)

3. ganeshie8

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

4. freckles

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

5. freckles

but this is correct to :p

6. freckles

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

7. 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

8. freckles

so you can skip finding A and B the icky way

9. ganeshie8

OMG! how triangular numbers popped up all off sudden!

10. freckles

I found the first few terms above

11. ganeshie8

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

12. freckles

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

13. Loser66

I am sorry. My computer is crazy!!

14. 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

15. ganeshie8

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

16. xapproachesinfinity

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

17. xapproachesinfinity

off*

18. xapproachesinfinity

Am i doing some kind of a mistake?

19. freckles

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

20. xapproachesinfinity

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

21. freckles

31*15

22. xapproachesinfinity

not on the choices

23. freckles

oh see what you are saying

24. xapproachesinfinity

i got the same thing

25. xapproachesinfinity

everything seems to me perfectly good given the pattern

26. freckles

27. freckles
28. freckles

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

29. xapproachesinfinity

oh ok

30. freckles

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

31. xapproachesinfinity

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

32. freckles

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

33. freckles

consecutive integers divided by 2 that is triangular right?

34. xapproachesinfinity

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

35. xapproachesinfinity

yeah seems to be it is

36. freckles

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

37. 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*

38. 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

39. freckles

hmmm...

40. freckles

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

41. xapproachesinfinity

hmm yeah i remember this

42. freckles

lol I don't remember

43. freckles
44. xapproachesinfinity

the algebra proof is the gauss's way :)

45. freckles

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

46. freckles

it does involve a bag of tricks

47. xapproachesinfinity

wow that GRE is not that easy haha

48. xapproachesinfinity

has a lot of questions that require some deep thinking lol

49. freckles

yep I don't remember it being that hard

50. xapproachesinfinity

you took one before?

51. freckles

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

52. freckles

53. xapproachesinfinity

i see! I have never taking any standardized test yet

54. freckles

wow lucky

55. xapproachesinfinity

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

56. xapproachesinfinity

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

57. xapproachesinfinity

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

58. freckles

Yeah. Very frustrating I bet.

59. 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 $a_n=\frac{n(n+1)}{2}$

60. 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*}

61. anonymous

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

62. Loser66

Wowwwwwwwwww. I need study harder!!!