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Thread: [SOLVED] Binomial Theorem(Sum of series)?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Binomial Theorem(Sum of series)?

    Find the sum of the series:
    1 + {(√2 - 1)/(2√2)} - {(2√2 - 3)/(12)} + {(5√2 - 7)/(48√2)} - {(12√2 - 17)/(480)} + ... upto ∞ terms.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Series

    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    Find the sum of the series:
    1 + {(√2 - 1)/(2√2)} - {(2√2 - 3)/(12)} + {(5√2 - 7)/(48√2)} - {(12√2 - 17)/(480)} + ... upto ∞ terms.
    Note that:

    • $\displaystyle (\sqrt2-1)^2 = (3-2\sqrt2)$


    • $\displaystyle (\sqrt2-1)^3 = (5\sqrt2-7)$


    • $\displaystyle (\sqrt2-1)^4 = (17 -12\sqrt2)$

    So, if $\displaystyle x = (\sqrt2-1)$ and we make all the signs positive, the numerators are $\displaystyle 1, x, x^2, x^3, x^4, ...$

    The denominators are: $\displaystyle 1, 1 \cdot2\sqrt2, 1 \cdot2\sqrt2\cdot 3\sqrt2,1 \cdot2\sqrt2\cdot3\sqrt2\cdot4\sqrt2,1 \cdot2\sqrt2\cdot3\sqrt2\cdot4\sqrt2\cdot5\sqrt2, ...$

    i.e. $\displaystyle 1, 2!\sqrt2, 3!(\sqrt2)^2,4!(\sqrt2)^3,5!(\sqrt2)^4, ...$

    So if we let $\displaystyle y =\frac{x}{\sqrt2}= \frac{\sqrt2-1}{\sqrt2}$ the series is

    $\displaystyle S=1 + \frac{y}{2!}+\frac{y^2}{3!}+\frac{y^3}{4!}+\frac{y ^4}{5!}+ ...$

    $\displaystyle \Rightarrow 1+ Sy = 1 +y+ \frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\frac {y^5}{5!}+ ...$

    = $\displaystyle e^y$

    $\displaystyle \Rightarrow S = \frac{e^y-1}{y}$, where $\displaystyle y = \frac{\sqrt2-1}{\sqrt2}$

    Grandad
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  3. #3
    Super Member fardeen_gen's Avatar
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    Thank you! That is indeed the answer. And I like your method very much.
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