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Math Help - Lost in proving .

  1. #1
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    Lost in proving .

    If S is the sum of the series 1+3x+5x^2+...+(2n+1)x^n , by considering (1-x)S , or otherwise , prove that .

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

    where x is not 1 .



    Thanks a lot .
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Hi

    I am gonna give you general proof for this AGP, Arithmetico-Geomtric Series

    a , (a+d) ,(a+2d)........ is an AP and b ,br,br^2....... is a GP

    The series we have is
    ab ,(a+d)br,(a+2d)br^2,.......

    Now
    S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}........(1)

    rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}..........(2)

    We now substract (1) -(2)

    (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n


    = \frac{ab}{1-r} + \frac{dbr(1-r^{n-1}}{(1-r)^2} - (a+(n-1)d)br^n

    \implies ~S_n = ~ \frac{ab}{1-r} + \frac{dbr(1-r^{n-1})}{(1-r)^2} - \frac{(a+(n-1)d)br^n}{1-r}
    .................................................. ...............................................

    Now you should always remember the way of proof of this thing for any general term and apply as it is in your case

    For going further , when -1< r <1


    Lim_{n\rightarrow \infty}~S_n = \frac{ab}{1-r} + \frac{dbr}{(1-r)^2}
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  3. #3
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    Quote Originally Posted by ADARSH View Post
    Hi

    Now
    S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}........(1)

    rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}..........(2)

    We now substract (1) -(2)

    (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n
    I can't understand this part . How did you get

    (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n

    when you subtract 2 from 1 .

    Thanks again .
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by ADARSH View Post
    H
    Now
    S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}........(1)

    rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}..........(2)

    We now substract (1) -(2)

    (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n

    ( 1) - (2) is given by

    (1-r)S_n = ab -abr +(a+d)br -abr^2+ (a+d)br^2 + ...... - (a+(n-1)d)br^n

    (1-r)S_n = ab~  +(a+d)br -abr + (a+d)br^2 -abr^2+ ...... - (a+(n-1)d)br^n

    See the things with -ve sign are from (2)

    Now open the brackets and go ahead to my step
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  5. #5
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    Quote Originally Posted by ADARSH View Post
    Hi



    \implies ~S_n = ~ \frac{ab}{1-r} + \frac{dbr(1-r^{n-1})}{(1-r)^2} - \frac{(a+(n-1)d)br^n}{1-r}
    .................................................. ...............................................

    THanks . If i plug these values in , then i will get my proof ?

    a=1
    b=1
    r=x
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    I have just given you a general proof to it.
    What you should do-

    - Don't just plug the values you won't be able to remember the formula for long

    -Try to remember the way it was proved and find the value of sum, for your a,b,r & d every time you get this type of question


    -If you do it this way , every question of A.G.P. will be solved
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