Question

Use mathematical induction to show that
1+2^1+2^2+⋯+2^n=2^(n+1)-1

Answer 1

is it true for n = 1

Answer 2

http://prntscr.com/57tcdd

Answer 3

@perl it's not a true or false, it's "show that"

Answer 4

yes, that is part of the induction proof though

Answer 5

'basis case'

Answer 6

we want to prove that
2^0+2^1+2^2+...+2^n=2^(n+1)-1 , is true for all n >= 0

basis case;
statement true for n=0 ?

inductive step:
if true for n=k, show it is true for n=k+1

Answer 7

ok so far?

Answer 8

Yes clear, I am listening

Answer 9

We want to prove that

2^0+2^1+2^2+...+2^n=2^(n+1)-1 , is true for all n >= 0

basis case;
statement true for n=0 ?

inductive step:
if true for n=k, show it is true for n=k+1



We want to prove that

2^0+2^1+2^2+...+2^n=2^(n+1)-1 , is true for all n >= 0

step 1: basis case;
statement true for n=0 ?
2^0 = 2^(0+1) -1
1 = 2-1
1=1
This is true (check).


Step 2: inductive case:

Prove 'if true for n=k, show it is true for n=k+1'

So we are given:

2^0 + 2^1 + ... + 2^k = 2^(k+1) - 1

add 2^(k+1) to both sides

2^0 + 2^1 + ... + 2^k + 2^(k+1)= 2^(k+1) - 1 + 2^(k+1)

(this is true since equals added to equals are equal)

Now simplify the right side of that , using properties of exponents.


2^0 + 2^1 + ... + 2^k + 2^(k+1)= 2^(k+1) + 2^(k+1) - 1


2^0 + 2^1 + ... + 2^k + 2^(k+1)= 2*2^(k+1) - 1


2^0 + 2^1 + ... + 2^k + 2^(k+1)= 2^(k+2) - 1

And this is what we wanted to prove, P(k) -> P(k+1)

Answer 10

@perl much appreciated!