Something is going on here. What would the equation be for 1003? What is the statement for n3? What about a proof?
Solution
First let us recognize that the strange numbers in the pattern, that is the numbers 1, 3, 6, 10 . . . are the sum of the first so many integers and that there is a formula for these, so-called triangular numbers, Tn:
T1T2T3T4T5T6======136101521======11+21+2+31+2+3+41+2+…+51+2+…+6======21⋅222⋅323⋅424⋅525⋅626⋅7.
If we take 53 as an example, we have
53=152−102=T52−T42=(T5+T4)(T5−T4)=(15+10)(15−10)=25⋅5=53
In general, we would like to have
n3=Tn2−Tn−12=(Tn+Tn−1)(Tn−Tn−1)
We can recall that Tn=1+2+…+n=n(n+1)/2.
Thus
Tn+Tn−1=2n(n+1)+2(n−1)n=2n2+n+(n2−n)=22n2=n2.
Similarly,
Tn−Tn−1=2n(n+1)−2(n−1)n=2n2+n−(n2−n)=22n=n.
So it works!
Tn2−Tn−12===(Tn−Tn−1)(Tn+Tn−1)n⋅n2n3.
Voila!