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Math Help - Proving Identities

  1. #1
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    Proving Identities

    Prove the following identities

    1*2^0 + 2*2^1 + 3*2^2 + . . . + n*2^2n-1= (n-1)2^n + 1


    Also

    1 + 3 + 9 + 27 + . . . + 3^n-1= (3n-1)
    2
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  2. #2
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    I assume the second one should be

    1 + 3 + 9 + 27 + \dots + 3^{n - 1} = \frac{3^n - 1}{2}.


    This is a geometric series with a = 1, r = 3.

    So S_n = \frac{a(r^n - 1)}{r - 1}

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

     = \frac{3^{n + 1} - 1}{2}.
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  3. #3
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    Hello, taylor1234!

    There is a typo in the first one . . . the general term is not correct.


    1\cdot2^0 + 2\cdot2^1 + 3\cdot2^2 + \hdots + \underbrace{n\cdot2^{2n-1}}_{???} \;=\; (n-1)2^n + 1

    In each term, the exponent is one less than the coefficient: . a_n \:=\:n\cdot 2^{n-1}


    We have: . . . . . S \;=\; 1\!\cdot\!2^0 + 2\!\cdot\!2^1 + 3\!\cdot\!2^2 + 4\!\cdot\!2^3 + \hdots + n\!\cdot\!2^{n-1}\qquad\qquad\qquad[1]

    Multiply by 2: . 2S \;=\; \qquad \quad 1\!\cdot\!2^1 + 2\!\cdot\12^2 + 3\!\cdot\!2^3 +   \hdots + (n-1)2^{n-1} + n\!\cdot\!2^n \;\;[2]<br />


    Subtract [2] - [1]: . S \;=\;-1 - 2 - 2^2 - 2^3 - \hdots - 2^{n-1} + n\cdot2^n

    . . . . . . . . . . . . . S \;=\;-\underbrace{(1 + 2 + 2^2 + 2^3 + \hdots 2^{n-1})}_{\text{geometric series}} +\; n\cdot 2^n

    . . . . . . . . . . . . . S \;=\;-\left(\dfrac{2^n-1}{2-1}\right) + n\cdot 2^n

    . . . . . . . . . . . . . S \;=\;-(2^n - 1) + n\cdot2^n

    . . . . . . . . . . . . . S \;=\;-2^n + 1 + n\cdot2^n

    . . . . . . . . . . . . . S \;=\;(n-1)2^n + 1
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  4. #4
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    How would I solve these problems using the induction method. Assuming Sn is true and testing for Sn+1?
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