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Math Help - Sum of a series

  1. #1
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    Sum of a series

    Given that S_n is the sum of a geometric series where S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}

    Find the values of a such that 256S_4=175S

    256S_4=175S

    256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175

    (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}

    \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}

    16(a-2)+4(a-2)^2+(a-2)^3=23.5
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Punch View Post
    Given that S_n is the sum of a geometric series where S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}

    Find the values of a such that 256S_4=175S

    256S_4=175S
    what is S??

    256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175
    how did you get 175+175 in the RHS??

    (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}

    \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}

    16(a-2)+4(a-2)^2+(a-2)^3=23.5
    ...
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  3. #3
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    Quote Originally Posted by Punch View Post
    Given that S_n is the sum of a geometric series where S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}
    I am going to assume that there should be "a- 2" in that last term.

    Find the values of a such that 256S_4=175S
    I am going to assume that "S" is the sum of the infinite series, which is \frac{4}{6-a}.

    256S_4=175S

    256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175
    How did 175S become 175+ 175? You should have
    256\left(1+ \frac{a-2}{4}+ \frac{(a- 2)^2}{16}+ \frac{a-2)^3}{64}\right)= \frac{700}{6- a}= \frac{700}{4- (a-2)}
    if you let y= a- 2 that becomes
    256\left(1+ \frac{y}{4}+ \frac{y^2}{16}+ \frac{y^3}{64}\right)= \frac{700}{4-y}
    \left(1+ \frac{y}{4}+ \frac{y^2}{16}+ \frac{y^3}{64}\right)= \frac{175}{64(4- y)}

    Multiplying on both sides by 64:
    64+ 16y+ 4y^2+ y^3= \frac{175}{4-y}

    Multiplying both sides by 4- y:
    256- y^4= 175
    so the whole problem reduces to
    y^4= 81


    (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}

    \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}

    16(a-2)+4(a-2)^2+(a-2)^3=23.5
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