Prove by mathematical induction that 1+3+5+7+....+(2n-1)=n²
n , n^2 , (2n-1) , sum
1 ,1, 1 ,1
2,, 4 ,3 4
3,, 9 ,5, 9
4 16 ,7 16 sure seems to work
1+3+5 .... = arithmetic series
a = 1
d = 2
but see https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
sum from k = 0 to k =n-1 of (a+kd) = (n/2)(2a+(n-1)d)
here:
(n/2)(2+ (n-1)2)
(n) (1 + (n-1))
(n) ( n)
n^2
1=1^2
assume true for n=k:
1+3+...+(2k-1) = k^2
Now check for n=k+1:
1+3+...+(2k-1)+(2k+1) = k^2+(2k+1) = (k+1)^2
So, P(k) => P(k+1)
P(1), so P(all k)
Go with Steve's way, I got off on series, not induction
To prove the given statement using mathematical induction, we need to show that it holds true for the base case (n = 1) and then prove the inductive step.
Step 1: Base Case (n = 1)
When n = 1, we have 1 + (2*1 - 1) = 1 + 1 = 2, and 1² = 1. Since 2 = 1², the statement holds true for the base case.
Step 2: Inductive Hypothesis
Assume that the statement holds true for some positive integer k. That is, assume that 1 + 3 + 5 + 7 + ... + (2k - 1) = k².
Step 3: Inductive Step
We need to prove that the statement holds true for k+1, assuming that it holds true for k.
For k+1, we have:
1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k+1) - 1)
Expanding the expression:
k² + (2(k+1) - 1)
k² + (2k + 2 - 1)
k² + 2k + 1
Factoring the expression:
(k+1)²
Since (k+1)² is equal to k² + 2k + 1, we can conclude that the statement holds true for k+1.
Step 4: Conclusion
By the principle of mathematical induction, we have proven that for all positive integers n, 1 + 3 + 5 + 7 + ... + (2n - 1) = n².