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    Hello!

    I do not know where to start on this proof by induction question

    The question is,

    Prove by mathematical induction that, for all positive integers n,

    1/2 + 1/4 + 1/8 + 1/16 ,,,, + 1/2^n = 1 - 1/2^n

    I just dont know where to start? I think method of differences as its a summation question but im not sure, btw how do i use proper mathematical notation on TSR? The way i've used '/' and '^' just looks confusing!

    Thank You,
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    (Original post by Jampolo)
    Hello!

    I do not know where to start on this proof by induction question

    The question is,

    Prove by mathematical induction that, for all positive integers n,

    1/2 + 1/4 + 1/8 + 1/16 ,,,, + 1/2^n = 1 - 1/2^n

    I just dont know where to start? I think method of differences as its a summation question but im not sure
    Consider the form of a proof by induction.

    Show a base case.

    Show that the case for k implies the case for k+1.


    You don't need the method of differences.

    btw how do i use proper mathematical notation on TSR? The way i've used '/' and '^' just looks confusing!

    Thank You,
    LaTex! See the reference at the top of the forum or top of this thread.
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    (Original post by ghostwalker)
    Consider the form of a proof by induction.

    Show a base case.

    Show that the case for k implies the case for k+1.


    You don't need the method of differences.



    LaTex! See the reference at the top of the forum or top of this thread.
    I cant show the the case for k+1, i dont know where to start, or more importantly, how to actually create and show a method of induction
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    ignore this post, i just cant use latex
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    Let n=1

    LHS = \frac{1}{2} RHS = 1-\frac{1}{2} = \frac{1}{2}

    Assume the result is true for n=k so that:
     \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^k} = 1 - \frac{1}{2^k}

    For n=k+1 I expect  1 - \frac{1}{2^{k+1}}

     \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^k} + \frac{1}{2^{k+1}} \equiv 1 - \frac{1}{2^k} + 1 - \frac{1}{2^{k+1}}

     1 - \frac{1}{2^k} + 1 - \frac{1}{2^{k+1}} = 2 - ( \frac{1}{2^k} + \frac{1}{2^{k+1}} )

    Then go on from there until you reach your expectation.

    Conclude by saying, 'So, if the result is true for n=k it is true for all n=k+1. As it is true for n=1 it is true for all n \geq 1 by induction.
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    Right i have an answer.. im not sure if its right because the answer at the end is different to what it should be, heck i dont even know if im taking the right steps

    Prove by mathematical induction that, for all positive integers n,

    1/2 + 1/4 + 1/8 + 1/16 ,,,, + 1/2^n = 1 - 1/2^n

    ok,

    1 \div 2 + 1\div 4 ..... + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}
    1 \div 2 + 1\div 4 ..... + \frac{1}{2^{k+1}} + \frac{1}{2^{k+1}} = 1
    1 \div 2 + 1\div 4 ..... + \frac{2}{2^{k+1}}= 1
    1 \div 2 + 1\div 4 ..... + \frac{2}{2 \times 2^k}= 1
    1 \div 2 + 1\div 4 ..... + \frac{1}{1\times 2^k}= 1
    1 \div 2 + 1\div 4 ..... + \frac{1}{2^k}= 1

    Hmm where do i stand with this?
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    (Original post by joshgoldman)
    Let n=1

    LHS = \frac{1}{2} RHS = 1-\frac{1}{2} = \frac{1}{2}

    Assume the result is true for n=k so that:
     \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^k} = 1 - \frac{1}{2^k}

    For n=k+1 I expect  1 - \frac{1}{2^{k+1}}

     \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^k} + \frac{1}{2^{k+1}} \equiv 1 - \frac{1}{2^k} + 1 - \frac{1}{2^{k+1}}

     1 - \frac{1}{2^k} + 1 - \frac{1}{2^{k+1}} = 2 - ( \frac{1}{2^k} + \frac{1}{2^{k+1}} )

    Then go on from there until you reach your expectation.

    Conclude by saying, 'So, if the result is true for n=k it is true for all n=k+1. As it is true for n=1 it is true for all n \geq 1 by induction.
    Ahh i get it, so is my post totally wrong? By your method i would 2=2 which obviously means that its true, but was i going in the right direction ?
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    (Original post by Jampolo)
    Ahh i get it, so is my post totally wrong? By your method i would 2=2 which obviously means that its true, but was i going in the right direction ?
    I'm not sure which exam board you're doing, but the one my school uses (OCR MEI) is very fussy even with the wording of the concluding sentence - you may want to check up on yours.
    It looks like you have re-arranged correctly in terms of a formula, but you must remember that this is a series and you are trying to prove something - rather than find the value of a variable.
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    (Original post by joshgoldman)
    I'm not sure which exam board you're doing, but the one my school uses (OCR MEI) is very fussy even with the wording of the concluding sentence - you may want to check up on yours.
    It looks like you have re-arranged correctly in terms of a formula, but you must remember that this is a series and you are trying to prove something - rather than find the value of a variable.
    oh yeah i would conclude, im on OCR (not mei) but i was just making sure the induction part was correct, the sum of that infinite series would be one, but im not just if thats enough of a justifacation to give me the marks through my method.
 
 
 
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