Challenge Level

Shubhangee from Buckler's Mead Academy in England and Sunhari from British School Muscat sent diagrammatic proofs. This is Shubhangee's proof:

Moncef from London Academy in Morocco constructed a similar but slightly different proof to the one in the proof sorter. Click here to see Moncef's proof.

Sunhari also submitted a proof by induction:

*Let P($n$) be the statement 'the sum of the first $n$ odd numbers is equal to $n^2$.*

P(1): 1 = 1^2

P(1) is true.

Suppose it is true for P($k$),

$1+3+… +(2k-1) = k^2$

Then, P($k+1$)

$1+3+…+[2(k+1) -1]

\\= k^2 + [2(k+1) -1]

\\= k^2 + 2k +2 -1

\\= k^2 + 2k + 1

\\= (k+1)^2$

*Therefore whenever P($k$) is true, P($k+1$) is also true. So since P(1) is true, P(2) must also be true, and so P(3) must also be true, and so on. This means P($n$) must be true for all positive integers $n$.*