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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

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  1. Loser66
    • one year ago
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    @Kainui @Michele_Laino

  2. freckles
    • one year ago
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    \[a_1=1 \\ a_2=\frac{3}{1}a_1=3(1)=3 \\ a_3=\frac{4}{2}a_2=2a_2=2(3)=6 \\ a_4=\frac{5}{3}a_3=\frac{5}{3}(6)=5(2)=10 \\ 1,3,6,10,... \text{ subtracting term from its previous term gives: } \\ 2,3,4,.. \\ \text{ doing that again we have } 1,1,1,... \\ \text{ so we have something \in the form } a_n=An^2+Bn+C \\ \text{ we could defined } a_0=0 \text{ since \it fits our pattern thingy } \\ \text{ so we could say } C=0 \\ a_n=An^2+Bn\] 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
    • one year ago
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    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
    • one year ago
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    \the thingy you had before was correct since a1 is 1

  5. freckles
    • one year ago
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    but this is correct to :p

  6. freckles
    • one year ago
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    I didn't think to do all of that that is pretty neat

  7. freckles
    • one year ago
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    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
    • one year ago
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    so you can skip finding A and B the icky way

  9. ganeshie8
    • one year ago
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    OMG! how triangular numbers popped up all off sudden!

  10. freckles
    • one year ago
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    I found the first few terms above

  11. ganeshie8
    • one year ago
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    Ahh i wasn't thinking.. \(\large \binom{n}{2}\) is a triangular number...

  12. freckles
    • one year ago
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    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
    • one year ago
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    I am sorry. My computer is crazy!!

  14. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    i got where ganesh got that formula \[a_{30}=\frac{31!}{2!29!}\] seems to me a bit of?

  17. xapproachesinfinity
    • one year ago
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    off*

  18. xapproachesinfinity
    • one year ago
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    Am i doing some kind of a mistake?

  19. freckles
    • one year ago
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    just need simplifying \[a_{30}=\frac{31 \cdot 30 \cdot 29!}{2 \cdot 29!}\]

  20. xapproachesinfinity
    • one year ago
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    yeah i did but does not look like it is giving the right answer

  21. freckles
    • one year ago
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    31*15

  22. xapproachesinfinity
    • one year ago
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    not on the choices

  23. freckles
    • one year ago
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    oh see what you are saying

  24. xapproachesinfinity
    • one year ago
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    i got the same thing

  25. xapproachesinfinity
    • one year ago
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    everything seems to me perfectly good given the pattern

  26. freckles
    • one year ago
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    oh loser made typeo

  27. freckles
    • one year ago
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    choice A is suppose to read (15)(31)

  28. xapproachesinfinity
    • one year ago
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    oh ok

  29. freckles
    • one year ago
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    yeah she is studying for the gre which she told me she was going to ace

  30. xapproachesinfinity
    • one year ago
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    i'm thinking or your question showing that (n+1)(n+2)/2 is triangular for n>0

  31. freckles
    • one year ago
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    (n+1)(n+2)/2 is triangular I thought

  32. freckles
    • one year ago
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    consecutive integers divided by 2 that is triangular right?

  33. xapproachesinfinity
    • one year ago
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    (n+1)(n+2)/2=(n+1)+n+......+1

  34. xapproachesinfinity
    • one year ago
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    yeah seems to be it is

  35. freckles
    • one year ago
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    ok you want to prove: \[\sum_{i=1}^{n}i=\frac{n(n+1)}{2} \\ \sum_{i=1}^{n+1}i=\frac{(n+1)(n+1+1)}{2}=\frac{(n+1)(n+2)}{2}\]

  36. freckles
    • one year ago
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    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*

  37. freckles
    • one year ago
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    \[S(n)=\sum_{i=1}^{n}i \\ S(n+1)=\sum_{i=1}^{n+1}i \\ S(n+1)-S(n)=(n+1)\] I think it starts something like this

  38. freckles
    • one year ago
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    hmmm...

  39. freckles
    • one year ago
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    could look it up but don't want to cheat yet

  40. xapproachesinfinity
    • one year ago
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    hmm yeah i remember this

  41. freckles
    • one year ago
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    lol I don't remember

  42. freckles
    • one year ago
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    http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

  43. xapproachesinfinity
    • one year ago
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    the algebra proof is the gauss's way :)

  44. freckles
    • one year ago
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    the proving the equality thing was easy by induction i just have a hard time remembering how to derive formulas like that

  45. freckles
    • one year ago
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    it does involve a bag of tricks

  46. xapproachesinfinity
    • one year ago
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    wow that GRE is not that easy haha

  47. xapproachesinfinity
    • one year ago
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    has a lot of questions that require some deep thinking lol

  48. freckles
    • one year ago
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    yep I don't remember it being that hard

  49. xapproachesinfinity
    • one year ago
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    you took one before?

  50. freckles
    • one year ago
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    there are questions on here that I'm not sure I can answer

  51. freckles
    • one year ago
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    yea about a decade ago

  52. xapproachesinfinity
    • one year ago
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    i see! I have never taking any standardized test yet

  53. freckles
    • one year ago
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    wow lucky

  54. xapproachesinfinity
    • one year ago
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    most of what is in that GRE is hard to me lol

  55. xapproachesinfinity
    • one year ago
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    what is the duration for such test? clearly this is not the exam it self?

  56. xapproachesinfinity
    • one year ago
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    it says 170 mins nearly 2 hour! for all that! that is a really frustrating hehe

  57. freckles
    • one year ago
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    Yeah. Very frustrating I bet.

  58. anonymous
    • one year ago
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    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 \[a_n=\frac{n(n+1)}{2}\]

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

  60. anonymous
    • one year ago
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    Oops, the generating function should be \(F(x)=\dfrac{2x-x^2}{\color{red}2(1-x)^2}\).

  61. Loser66
    • one year ago
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    Wowwwwwwwwww. I need study harder!!!

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