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    (Original post by twig)
    Nice question, where did you find it?
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     \frac{ 1}{1-r}( \frac{ r(1-{r}^{n} )}{1-r}-n {r}^{n+1}  ) , using double summations of G.P.s which varied in length in an A.P. Is the result correct? And what's the other way?


    Regarding USA, know of any other exchange programs around? I think Cambridge does one with MIT, but that's all I know...
    \displaystyle\sum_{i=1}^nir^i = r*(\displaystyle\sum_{i=1}^nir^{  i-1}), and then we can sum what's in the brackets by integrating term by term (and then comparing both sides using a nice value of r to find the constant of integration).
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    (Original post by around)
    \displaystyle\sum_{i=1}^nir^i = r*(\displaystyle\sum_{i=1}^nir^{  i-1}), and then we can sum what's in the brackets by integrating term by term (and then comparing both sides using a nice value of r to find the constant of integration).
    Yeah, r \frac{ d}{dr }( \frac{ r(1-r^{n}}{1-r } ) gives the same expression, using limit r and 0 from your integral.


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    I think another easy way is just to consider how we derive the normal sum of G.P., with little modifications. THis maybe the second C1/c2 method ben implied?
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    (Original post by twig)
    Nice question, where did you find it?
    Spoiler:
    Show
     \frac{ 1}{1-r}( \frac{ r(1-{r}^{n} )}{1-r}-n {r}^{n+1}  ) , using double summations of G.P.s which varied in length in an A.P. Is the result correct? And what's the other way?


    Regarding USA, know of any other exchange programs around? I think Cambridge does one with MIT, but that's all I know...


    EDIT: Have seen the other way now.
    Something like that. As I said, it pops up alot in STEP probability questions. with regards to the USA, I don't know of many:
    -CAM-MIT
    -OX-Princeton (I think)
    -Leicester does one


    (Original post by around)
    \displaystyle\sum_{i=1}^nir^i = r*(\displaystyle\sum_{i=1}^nir^{  i-1}), and then we can sum what's in the brackets by integrating term by term (and then comparing both sides using a nice value of r to find the constant of integration).
    wasn't quite what I was thinking...
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    \sum_{i=1}^nir^i=r*(\sum_{i=1}^n  ir^{i-1})=r*(\sum_{i=1}^n\frac{d}{dr}[r^{i}]) differentiation is a linear operator so you can swap the sigma and the derivative operator round and bob's your uncle (or however the saying goes)


    EDIT: sorry twig, didn't see your post.
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    (Original post by ben-smith)

    wasn't quite what I was thinking...
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    \sum_{i=1}^nir^i=r*(\sum_{i=1}^n  ir^{i-1})=r*(\sum_{i=1}^n\frac{d}{dr}[r^{i}]) differentiation is a linear operator so you can swap the sigma and the derivative operator round and bob's your uncle (or however the saying goes)


    EDIT: sorry twig, didn't see your post.
    We wrote exactly the same thing...

    EDIT: I think the Cam-MIT exchange was cut due to lack of funding (or something similar). Also, they weren't too keen on mathematicians doing it because apparently MIT 3rd year was just too easy for Cambridge maths undergrads.
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    C1 > 94
    C2 >100
    C3 > 96
    M1 > 85
    D1 > 89
    S1 > 97

    Finishing the rest of maths and f.maths next year

    I also do chem and physics but probs gonna drop physics next year
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    (Original post by twig)
    Yeah, r \frac{ d}{dr }( \frac{ r(1-r^{n}}{1-r } ) gives the same expression, using limit r and 0 from your integral.


    Spoiler:
    Show
    I think another easy way is just to consider how we derive the normal sum of G.P., with little modifications. THis maybe the second C1/c2 method ben implied?
    your first approach and the one I just posted are the ones I was talking about. The first is my favourite as it is the way I first did it and I worked out simply by writing the terms out in a grid. I then summed down as opposed to sideways which makes it a sum of a sums of geometric series. I like geometrical arguments.
    The second approach seems to work but it makes me uneasy as I have a sneaky feeling that it shouldn't strictly be allowed. I don't know why.
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    (Original post by ben-smith)
    your first approach and the one I just posted are the ones I was talking about. The first is my favourite as it is the way I first did it and I worked out simply by writing the terms out in a grid. I then summed down as opposed to sideways which makes it a sum of a sums of geometric series. I like geometrical arguments.
    The second approach seems to work but it makes me uneasy as I have a sneaky feeling that it shouldn't strictly be allowed. I don't know why.
    Everything converges absolutely so term by term differentiation/integration is ok. Good that you're thinking about it though.
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    (Original post by around)
    We wrote exactly the same thing...

    EDIT: I think the Cam-MIT exchange was cut due to lack of funding (or something similar). Also, they weren't too keen on mathematicians doing it because apparently MIT 3rd year was just too easy for Cambridge maths undergrads.
    oops. Many apologies. You mentioned integrating and what I had in mind involved differentiating so I assumed we were talking about different things. I see now. I hardly think I'll get into Cam at this rate

    That's a pity about the exchange. It sounded very enjoyable.
    What courses you taking this coming year?
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    (Original post by ben-smith)
    oops. Many apologies. You mentioned integrating and what I had in mind involved differentiating so I assumed we were talking about different things. I see now. I hardly think I'll get into Cam at this rate

    That's a pity about the exchange. It sounded very enjoyable.
    What courses you taking this coming year?
    Too many, now I come to think about it. I'm mainly taking a mix of algebra and geometry, with some set theory thrown in. I've started to get more and more interested in foundations recently.
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    (Original post by around)
    Too many, now I come to think about it. I'm mainly taking a mix of algebra and geometry, with some set theory thrown in. I've started to get more and more interested in foundations recently.
    Foundations? Stuff like Godel's incompleteness theorem, lambda calculus? That's cool. I've been reading about it in Alan Turing's biography.
    No Physics? Not tempted by GR?
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    (Original post by ben-smith)
    your first approach and the one I just posted are the ones I was talking about. The first is my favourite as it is the way I first did it and I worked out simply by writing the terms out in a grid. I then summed down as opposed to sideways which makes it a sum of a sums of geometric series. I like geometrical arguments.
    The second approach seems to work but it makes me uneasy as I have a sneaky feeling that it shouldn't strictly be allowed. I don't know why.
    Regarding the first method: I first expanded the sum, then rewrote it as, which is the same thing I guess:

    \sum_{j=1}^n\sum_{i=j}^n r^{i}
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    (Original post by ben-smith)
    Foundations? Stuff like Godel's incompleteness theorem, lambda calculus? That's cool. I've been reading about it in Alan Turing's biography.
    No Physics? Not tempted by GR?
    GR is basically geometry ().
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    (Original post by twig)
    Yeah, I first expanded the sum, then rewrote it as, which is the same thing I guess:

    \sum_{j=1}^n\sum_{i=j}^n r^{i}
    your turn to ask me a question
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    Surely solving that using differentiation and integration is not the C2 way?
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    (Original post by deejayy)
    Surely solving that using differentiation and integration is not the C2 way?
    Oh yeah. It's been too long since I did any C2.
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    (Original post by ben-smith)
    Oh yeah. It's been too long since I did any C2.
    What module is it where you use integration to solve that?
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    (Original post by deejayy)
    Surely solving that using differentiation and integration is not the C2 way?
    Integration was in my C2...
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    (Original post by ben-smith)
    For example, how would you go about finding:
    \displaystyle\sum_{i=1}^nir^i?
    \sigma(n) := \sum_{k=0}^nkx^k = \sum_{k=-1}^{n-1}(k+1)x^{k+1}  = -(n+1)x^{n+1}+\sum_{k=0}^{n}(k+1)  x^{k+1}  = -(n+1)x^{n+1}+x\sigma(n)+\frac{x(  x^{n+1}-1)}{x-1}.
    Simple rearranging now gives you the answer of \sigma(n) = \frac{ (n x-n-1) x^{n+1}+x}{(1-x)^2}. Best of luck everyone, by the way, with applications and everything.
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    (Original post by deejayy)
    What module is it where you use integration to solve that?
    C4 I guess but you can do the same thing by differentiating with C3 techniques.
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    (Original post by L'art pour l'art)
    \sigma(n) := \sum_{k=0}^nx^k = \sum_{k=-1}^{n-1}(k+1)x^{k+1}  = -(n+1)x^{n+1}+\sum_{k=0}^{n}(k+1)  x^{k+1}  = -(n+1)x^{n+1}+x\sigma(n)+\frac{x(  x^{n+1}-1)}{x-1}.
    Simple rearranging now gives you the answer of \sigma(n) = \frac{ (n x-n-1) x^{n+1}+x}{(1-x)^2}. Best of luck everyone, by the way, with applications and everything.
    Nice, a third method!

    EDIT:Cheers for this. I'd never even thought of this.
 
 
 
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