For all integers n ≥ 1, prove the following statement using mathematical induction. 1+2^1 +2^2 +...+2^n = 2^(n+1) −1
Here's what I have so far
1. Prove the base step
let n=1
2^1=2^(1+1)-1
False.
Someone else suggested that the base step is :
1+2^1=2^(1+1)-1
This returns a true value of 3=3 but doesn't work for any other n... Can some one tell me what the base step is? and why?
2. State the inductive hypothesis.
Assume the statement is also true for k
3. State what you have to show.
4. Proof proper:
sure it works.
1 = 2^1-1
1+2 = 2^2-1
1+2+4 = 2^3-1
...
So, the base step works fine.
Assume it's true for n=k. Now, if n=k+1 we have
1+2+4+...+2^k+2^(k+1) = 2^(k+1)-1 + 2^(k+1)
= 2*2^(k+1)-1
= 2^(k+2)-1
= 2^((k+1)+1)-1
QED
2+4+6+....+2n=n(n+1)
The base step for this proof is to show that the statement holds true for n=1.
Let n=1:
1+2^1 = 2^(1+1) - 1
1+2 = 4 -1
3 = 3
Since both sides of the equation are equal, the statement holds true for n=1.
Now, let's move on to the inductive hypothesis.
Assume that the statement is true for some integer k ≥ 1:
1+2^1 + 2^2 + ... + 2^k = 2^(k+1) - 1
We need to show that the statement is also true for k+1.
Starting with the left-hand side of the equation:
1+2^1 + 2^2 + ... + 2^k + 2^(k+1)
Using the inductive hypothesis, we can substitute the expression for k:
2^(k+1) - 1 + 2^(k+1)
Now, combine like terms:
2^(k+1) + 2^(k+1) - 1
Using the properties of exponents, we can simplify:
2 * 2^(k+1) - 1
Applying the exponent rule:
2^(k+2) - 1
Which is equivalent to the right-hand side of the original statement:
2^(k+1+1) - 1
Thus, by mathematical induction, we can conclude that the statement holds true for all integers n ≥ 1.
To prove the statement using mathematical induction, we need to go through four steps: the base step, the inductive hypothesis, the inductive step, and the conclusion.
1. Base Step:
The base step requires us to prove the statement for the smallest possible value of n, which is n=1 in this case.
We want to show that 1 + 2^1 + 2^2 + ... + 2^1 = 2^(1+1) - 1.
When n=1:
1 + 2^1 = 2^2 - 1
3 = 3
The base step is indeed true, as 3=3. Therefore, the statement holds for n=1.
2. Inductive Hypothesis:
Assume that the statement is true for some positive integer k, which means that
1 + 2^1 + 2^2 + ... + 2^k = 2^(k+1) - 1.
3. Inductive Step:
We need to prove that if the statement is true for k, then it is also true for k+1.
Consider the sum 1 + 2^1 + 2^2 + ... + 2^k + 2^(k+1).
Using the inductive hypothesis, we substitute the sum of the first k terms:
1 + 2^1 + 2^2 + ... + 2^k + 2^(k+1) = (2^(k+1) - 1) + 2^(k+1).
Now we simplify the right-hand side:
(2^(k+1) - 1) + 2^(k+1) = 2^(k+1) - 1 + 2^(k+1) = 2 * 2^(k+1) - 1 = 2^(k+2) - 1.
Thus, we have shown that if the statement holds for k, it also holds for k+1.
4. Conclusion:
By mathematical induction, we have proven that the statement is true for all integers n ≥ 1.
To summarize the proof:
Base Step: The statement holds true for n=1.
Inductive Hypothesis: Assume the statement is true for k.
Inductive Step: Prove that if the statement is true for k, it is also true for k+1.
Conclusion: By mathematical induction, the statement holds true for all integers n ≥ 1.