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    Hello all,

    I have been asked to read 'mathematical techniques' in order to prepare for my studies next year.
    I am currently at a chapter about sums to infinity and I have never seen this before (I studied in Belgium)

    The 2 pages of explanation are not too hard, but at the end there is a self-test that I just can't solve :woo:

    The question is:
    Sum the series 1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^N.
    What is the sum to infinity of the series if |x| < 1?

    I am quite clueless... the answer is supposed to be 3/(1-x)² - 2/(1-x)

    Any help appreciated, not too complicated though :p:
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    Rewrite it using sum notation and then split it into two sums. Then do the sum of infinity.

    Its hard because what do you know?

    \sum_1^{\infty}3nx^{n}+\sum_1^{\  infty}(-2)x^{n}
    Do you know what to do from here?
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    So you need to get it in the form of a geometric progression.

    To do that you're going to need to eliminate the arithemetic progression in the "constant" terms.

    There is a standard technique of multiplying the series by the geometric term (in this case x) and subtracting it from the original expression. This will give you something that looks more like a G.P. Have a go.
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    ghostwalker I was under the impression you just needed to integrate and differentiate 3nx^n by something. Although, I haven't done any problems like this.
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    (Original post by Simplicity)
    ghostwalker I was under the impression you just needed to integrate and differentiate 3nx^n by something. Although, I haven't done any problems like this.
    Not sure where you're coming from with that. Can you elaborate?
    Or point me at a webpage.

    Edit: Yeah. Uni level though? Don't know what level the OP is at; can't tell from their profile.

    Edit and again: Just realised it is a finite sum taken to the limit, so yeah could be A level. Don't know what they teach nowadays.
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    Integrate term by term to get rid of the n, find out an expression for the sum then differentiate that to get the sum of your original series.
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    Thanks for the quick replies already, although it's still not quite clear to me :s:

    I just finished secondary education, I know how to integrate, differentiate etc, but this is completely new to me and the book regards it as 'revision' so there is not that much explanation.

    I tried the 3 solutions posted above and always get stuck somewhere, or find a completely different solution I must be doing something wrong

    If someone could post it step by step it might dawn on me, or perhaps point me in the direction of a helpful site that explains it thoroughly, I used google but didn't find a good site.

    Thanks, Laurens
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    Are you sure its not 1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^(N - 1) because that would make it easier.

    Yes, surely it has to be x^(N-1) so the terms would actually come out.
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    (Original post by ShortRef)
    Are you sure its not 1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^(N - 1) because that would make it easier.

    Yes, surely it has to be x^(N-1) so the terms would actually come out.
    Good point. Otherwise the terms don't evaluate correctly if you set N to a small value.
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    Something which wil work:

    S = 1 + 4x + 7x^2 + ...
    Sx = x + 4x^2 +7x^3 + ... (we have multiplied every term by x here)

    S - Sx = 1 + 3x + 3x^2 + ...

    S(1-x) = 1 + 3x/(1-x) (we spot that there's a geometric series and sum accordingly)

    S = (1+2x)/(1-x)^2

    Sorry for misleading you first time round.
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    (Original post by around)
    As Simplicity said, you can split up your sum into two separate sums:

    \displaystyle\sum_{n=1}^{\infty} (3n-2)x^n = \displaystyle\sum_{n=1}^{\infty} 3nx^n -\displaystyle\sum_{n=1}^{\infty} 2x^n

    The second sum is easy enough to work out, but to deal with the first sum, integrate term by term:

    \displaystyle\int\sum_{n=1}^{\in  fty} 3nx^n\ dx = \displaystyle\sum_{n=1}^{\infty} 3x^{n+1}

    You should be able to work out the sum of the integrated series, and so to get the sum of your original series you have to differentiate your expression.

    Does this help?
    I dont think that would work, because you still need the formula for the sum of the series for nx^(n), which you dont have.

    If it happens to be x^(n-1) start with series of x^n (and with the sum of its infinite series) and differentiate both sides. If not, then I dont know, unless im wrong about the above post ^^^^^^ being wrong.
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    (Original post by around)
    As Simplicity said, you can split up your sum into two separate sums:

    \displaystyle\sum_{n=1}^{\infty} (3n-2)x^n = \displaystyle\sum_{n=1}^{\infty} 3nx^n -\displaystyle\sum_{n=1}^{\infty} 2x^n

    The second sum is easy enough to work out, but to deal with the first sum, integrate term by term:

    \displaystyle\int\sum_{n=1}^{\in  fty} 3nx^n\ dx = \displaystyle\sum_{n=1}^{\infty} 3x^{n+1}

    You should be able to work out the sum of the integrated series, and so to get the sum of your original series you have to differentiate your expression.

    Does this help?
    Call me stupid. But, the integration is wrong. You have to divide by n+1 and add C.
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    whoops. i am indeed wrong about the integration.

    i apologise to everyone in advance, and the post has now been amended to an approach that does work.

    ps you don't need the +c seeing as you differentiate again anyway.
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    (Original post by ShortRef)
    Are you sure its not 1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^(N - 1) because that would make it easier.

    Yes, surely it has to be x^(N-1) so the terms would actually come out.
    Well I noticed this as well, but the book really says ^N so I thought I was wrong, then again it wouldn't be the first mistake I find in this book, some spelling errors here and there :p:

    Thanks again for the quick replies guys, gonna go up to my room and see if I can do it with the new methods posted!
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    Without integrating (and barring errors on my part):

    Spoiler:
    Show


    \text{Let } S_N=1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^{(N - 1)}

    \text{Then } xS_N=x + 4x² + 7x³ + ... + (3N - 5)x^{(N - 1)}+(3N - 2)x^N

    Subtracting:

    (1-x)S_N=1+3x + 3x² + 3x³ + ... + 3x^{(N - 1)}-(3N - 2)x^N

    =3+3x + 3x² + 3x³ + ... + 3x^{(N - 1)}-(3N - 2)x^N-2

    =3(1+x + x² + x³ + ... + x^{(N - 1)})-(3N - 2)x^N-2

    =3(\frac{1-x^n}{1-x})-(3N - 2)x^N-2

    So

    S_N=3(\frac{1-x^n}{(1-x)^2})-\frac{(3N - 2)x^N}{1-x}-\frac{2}{1-x}

    And take the limit as n tends to infinity.



    Somewhat more cumbersome than the integral/differentiation method, I have to admit.
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    (Original post by around)
    Something which wil work:

    S = 1 + 4x + 7x^2 + ...
    Sx = x + 4x^2 +7x^3 + ... (we have multiplied every term by x here)

    S - Sx = 1 + 3x + 3x^2 + ...

    S(1-x) = 1 + 3x/(1-x) (we spot that there's a geometric series and sum accordingly)

    S = (1+2x)/(1-x)^2

    Sorry for misleading you first time round.
    Did you use x^n or x^(n-1)....its kind of a good idea to leave the last term in terms of x and n in a series :>
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    x^n, also didn't see that they wanted a finite sum first, sorry.

    i'm not having a good maths day today :<
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    I used x^n and it works.

    Seriously, the book is correct you guy are wrong. Also, if you do change it the last term doesn't make sense.

    s=1+4x^2+....+(3n-5)x^{n-1}+(3n-2)x^n

    (1-x)s=3+3x+....+3x^n - (3n-2)x^{n+1}-2

    Then its pretty much the same method thing used.

    I don't know why you did that as it makes it wrong. because n-1 term must be 3(n-1)-2 to make sense not 3n-2. Plus, it works like this.
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    Can we please go back and actually discuss if the question is right or not.

    "Sum the series 1 + 4x + 7x² + 10x³ + ... + (3N - 2)x^N."

    so the general term is (3N - 2)x^N

    Putting N = 0 gives -2
    Putting N = 1 gives (3 - 2)x^1 = x
    Putting N = 2 gives (6 - 2)x^2 = 4x^2
    Putting N = 2 gives (9 - 2)x^3 = 7x^3

    So the series has to be (3n - 2)x^(n-1)

    (Original post by Simplicity)

    I don't know why you did that as it makes it wrong. because n-1 term must be 3(n-1)-2 to make sense not 3n-2. Plus, it works like this.
    We dont replace n with n - 1, we change the power to n - 1.
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    (Original post by ShortRef)
    So the series has to be (3n - 2)x^(n-1)
    No that is wrong. When did it say n=0. I'm pretty sure it starts at n=1 and goes on.

    Again, your wrong it works fine without replacing n with n-1.
 
 
 
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