Prove by mathematical induction that 1+3+5+7+......+(2n-1)=n²
Proof by induction takes two steps.
Step 1: Show that the left hand side equals the right hand side for n = 1
LS... 2(1) - 1 = 1
RS... 1^2 = 1
So step 1 holds true, thus meaning we can proceed to step 2 the induction step:
Assume true for n=k and show true for n=k+1
Show that the LS = RS
1+ 3+5... + (2k-1) + {2(k+1) - 1} = (k+1)^2
Now we know that 1+3+... (2k-1) is equal to k^2 so replace this one the left side
k^2 + {2(k+1) - 1} = (k+1)(k+1)
Now simplify both sides by expanding...
You will see that the LS = RS thus 1+ 3 + 5 +... = n^2
To prove the given statement using mathematical induction, we need to first verify the base case. Then we assume that the statement holds for some value 'k', and prove that it also holds for 'k+1'.
Step 1: Base Case
Let's first verify the base case where n = 1:
When n = 1, the left-hand side of the equation becomes:
1 = 1²
Since the left and right sides are equal, the statement holds for n = 1.
Step 2: Inductive Assumption
Assume that the statement holds for some positive integer 'k':
1 + 3 + 5 + 7 + ... + (2k-1) = k²
Step 3: Inductive Step
We need to prove that the statement holds for (k + 1).
Consider the left-hand side of the equation when n = k + 1:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1)
By expanding the expression, we get:
k² + 2k + 1
Now, let's simplify the expression:
k² + 2k + 1 = (k + 1)²
Since the left and right sides are equal, the statement holds for n = k + 1.
Therefore, by the principle of mathematical induction, the given statement is proved to be true for all positive integers n.
To prove the given statement by mathematical induction, we'll follow the standard approach consisting of two steps:
Step 1: Base Case
Step 2: Inductive Case
Step 1: Base Case
Let's start by verifying the statement for n = 1:
1 + (2*1 - 1) = 1^2
1 + 1 = 1, which is true.
Step 2: Inductive Case
Assuming the statement is true for n = k (where k is an arbitrary positive integer), we will prove it for n = k + 1.
Now, let's assume the given statement holds for n = k:
1 + 3 + 5 + 7 + ... + (2k - 1) = k^2 (Inductive Hypothesis)
Now, we need to show that it holds for n = k + 1:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = (k + 1)^2
Adding (2(k + 1) - 1) to both sides of the equation:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = k^2 + (2(k + 1) - 1)
Simplifying the right-hand side:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = k^2 + (2k + 2 - 1)
Combining like terms:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = k^2 + 2k + 1
Factoring the right-hand side:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1) = (k + 1)^2
This completes the inductive step. Therefore, by the principle of mathematical induction, we have proven that 1 + 3 + 5 + 7 + ... + (2n - 1) = n^2 for all positive integers n.