How to do this following "Induction sum"?

n

SUM (-1/2)^j = (2^n+1 + (-1)^n)/ 3. 2^n

j=0

How do we remove the "summation" sign here? Also, how to go about proving it by using "induction"?

Thanks for you reply!

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- May 30th 2008, 09:24 AMrobocop_911Prove by Induction
How to do this following "Induction sum"?

n

SUM (-1/2)^j = (2^n+1 + (-1)^n)/ 3. 2^n

j=0

How do we remove the "summation" sign here? Also, how to go about proving it by using "induction"?

Thanks for you reply! - May 30th 2008, 09:38 AMflyingsquirrel
Hello

To prove this by induction, check that the equality is true for . (easy :D)

Then, assuming the equality is true for an integer ,

Multiply the first fraction by and the second one by so that they can be added.

Does it help ? - May 30th 2008, 09:54 AMrobocop_911
Hi, Thanks for you quick reply man!

Anyways, I didn't got how did you "expand"

n

sum (-1/2)^j into (2^n+1+(-1)^n)/3.2^n etc...

j =0

Also, how to get the symbols like you are using? Do we have to write the code for that?

Thanks - May 30th 2008, 10:18 AMflyingsquirrel
I assumed the equality was true for . :D

Quote:

Also, how to get the symbols like you are using? Do we have to write the code for that?

- May 30th 2008, 10:43 AMrobocop_911
- May 30th 2008, 12:16 PMflyingsquirrel
I did this so that the two fractions share the same denominator.

As , we choose as the common denominator. To make appear at the denominator of the first fraction, we multiply it by . For a similar reason, we multiply the second fraction by .

Otherwise, one can do this as follows :