Prove by mathematical induction that 1+3+5+7+....+(2n-1)=n²

Answer 1

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)

Answer 2

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

Answer 3

Go with Steve's way, I got off on series, not induction