Question

prove by mathematical induction 1+2+2^2+.......+2^n-1=2^n-1

Answer 1

how does 2^n-1 = 2^n-1 + (alot) ??

Answer 2

maybe its meant to read as:
\[...+2^{n-1}=2^n-1\]

Answer 3

yep yep.

Answer 4

provide a basis step; is it true for n=1?

if so, assume its true for any k value of n, k e Z, or do we need to expand the domain?

Answer 5

i dont noe

Answer 6

well, without knowing what you are working with in class, its hard to give an accurate solution. We can most likely do this for the domain of integers, but if the question is asking for all real numbers than the "k e Z" is not proper

Answer 7

well we have to prove that it is true for all real n

Answer 8

1+2+2^2+.......+2^n-1 hmmm, from sequence of exponents, it looks like its restricted to just the non negative integers; such that n= {1,2,3,4,...}

Answer 9

the correction you made above is correct i just did not know how to type it in

Answer 10

after proving its true for n=1, assume it is true for n=k

1 + 2 + 2^2 + ... + 2^(k-1) = 2^k - 1

and add 2^((k+1)-1) to each side to prove it for the (k+1)th value

\[1 + 2 + 2^2 + ... + 2^{k-1} + 2^{(k+1)-1} = 2^k - 1 + 2^{(k+1)-1}\]

Answer 11

now, keep in mind that we are trying to get to the form "\(=2^{k+1}-1\)" for the proof

Answer 12

\[1 + 2 + 2^2 + ... + 2^{k-1} + 2^{(k+1)-1} = 2^k - 1 + 2^{(k+1)-1}\]
\[1 + 2 + 2^2 + ... + 2^{k-1} + 2^{k} = 2*2^k - 1\]
\[1 + 2 + 2^2 + ... + 2^{k-1} + 2^{k} = 2^{k+1} - 1\]
... maybe, these still give me issues :)