You may also like

Latin Numbers

Can you create a Latin Square from multiples of a six digit number?

Summing Consecutive Numbers

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?

Number Sandwiches

Can you arrange the digits 1, 1, 2, 2, 3 and 3 to make a Number Sandwich?

Adding Odd Numbers

Age 11 to 16
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 $n$.