IFF - Interactive Proof Sorter

Stage: 4 and 5 Challenge Level: Challenge Level:1
something something sorted not sorted

Then \(T=\frac12n(n+1)\) for some whole number \(n\)

Expanding, \(8T+1=4n^2+4n+1\)

Let \(T\) be a triangular number

Therefore, if \(T\) is triangular, \(8T+1\) is square

We wish to prove that if \(T\) is a triangular number then \(8T+1\) is a square number.

Simplifying, \(8T+1=4n(n+1)+1\)

Factorising the right hand side, \(8T+1=(2n+1)^2\)

Therefore \(8T+1 = 8(\frac12n(n+1))+1\)