Mathematical Induction

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Johnny Blaze

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Hi everyone...I need a help. Please prove by the mathematical induction method for the question below:

1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ n/2^n = 2 - (n+2/2^n)

Thank you!
 
Johnny Blaze said:
Hi everyone...I need a help. Please prove by the mathematical induction method for the question below:

1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ n/2^n = 2 - (n+2/2^n)

Thank you!
Click to expand...
Can you please tell us the steps of "mathematical induction"?
 
As our name implies, we give help rather than answers. So you need to show us what you have done (even if you know it is wrong) so that we can see EXACTLY WHERE you are getting confused. Sometimes, you do not have any work to show because you did not even know where to start. In that case, it is best to start with basic definitions as Subhotosh Khan has suggested. With proofs by induction, it is also frequently useful when there are terms that follow a repeated pattern to create a general expression for that pattern.

So I shall provide a hint that will be useful after you answer SK's question.

The general form of those fractions is

[MATH]\dfrac{m}{2^m}, \text { where } m \text { is a positive integer.}[/MATH]
 
First of all, the way you have written this,
1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ n/2^n = 2 - (n+2/2^n)
Click to expand...
It is NOT TRUE! As you would have seen when you did the "first step":
when n= 1 the left side is 1/2 but the right side is 2- (1+ 2/2)= 0.

You must mean
1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ n/2^n = 2 - (n+2)/2^n
Click to expand...
Now both left and right sides are 1/2 when n= 1.

For n= k, we have
1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ k/2^k = 2 - (k+2)/2^k.
Setting n= k+ 1 just adds a new term to the left side and so also to the right side:
1/2 + 2/4 + 3/8 + 4/16 + 5/32 +...+ k/2^k+ (k+1)/2^{k+1) = 2 - (k+2)/2^k+ (k+1)/2^{k+1}

Do the algebra on the right. Is it the same as "2- (k+1+2)/2^{k+1}"?
 
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