1730.32 – Squares and Cubes in a Pattern

Here is the pattern:

13=12−0223=32−1233=62−3243=102−6253=152−10263=212−152 \begin{aligned} 1^3 &=& 1^2 - 0^2 \\ 2^3 &=& 3^2 - 1^2 \\ 3^3 &=& 6^2 - 3^2 \\ 4^3 &=& 10^2 - 6^2 \\ 5^3 &=& 15^2 - 10^2 \\ 6^3 &=& 21^2 - 15^2 \end{aligned}

Something is going on here. What would the equation be for 1003100^3? What is the statement for n3n^3? 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, TnT_n:

T1=1=1=1⋅22T2=3=1+2=2⋅32T3=6=1+2+3=3⋅42T4=10=1+2+3+4=4⋅52T5=15=1+2+…+5=5⋅62T6=21=1+2+…+6=6⋅72. \begin{aligned} T_1 &=& 1 &=& 1 &=&\frac{1 \cdot 2}{2} \\ T_2 &=& 3 &=& 1 + 2 &=&\frac{2 \cdot 3}{2} \\ T_3 &=& 6 &=& 1 + 2 + 3 &=&\frac{3 \cdot 4}{2} \\ T_4 &=& 10 &=& 1+2+3+4 &=& \frac{4 \cdot 5}{2} \\ T_5 &=& 15 &=& 1+2+ \ldots +5 &=& \frac{5 \cdot 6}{2} \\ T_6 &=& 21 &=& 1 +2 + \ldots +6 &=&\frac{6 \cdot 7}{2}. \\ \end{aligned}

If we take 535^3 as an example, we have

53=152−102=T52−T42=(T5+T4)(T5−T4)=(15+10)(15−10)=25⋅5=53 \begin{aligned}5^3 = 15^2 - 10^2 = T_5^2 - T_4^2 = (T_5 + T_4)(T_5 - T_4) = (15 + 10)(15 - 10) = 25 \cdot5 = 5^3\end{aligned}

In general, we would like to have

n3=Tn2−Tn−12=(Tn+Tn−1)(Tn−Tn−1) \begin{aligned}n^3 = T_n^2 - T_n-1^2 = (T_n + T_n-1)(T_n - T_n-1)\end{aligned}

We can recall that Tn=1+2+…+n=n(n+1)/2. T_n = 1 + 2 + \ldots + n = n (n + 1)/2.

Thus

Tn+Tn−1=n(n+1)2+(n−1)n2=n2+n+(n2−n)2=2n22=n2.T_n + T_{n-1} = \frac{n (n+1)}{2} + \frac{(n - 1) n}{2} = \frac{n^2 + n + (n^2 - n)}{2} = \frac{2 n^2}{2} = n^2.

Similarly,

Tn−Tn−1=n(n+1)2−(n−1)n2=n2+n−(n2−n)2=2n2=n.T_n - T_{n-1} = \frac{n (n+1)}{2} - \frac{(n - 1) n}{2} = \frac{n^2 + n - (n^2 - n)}{2} = \frac{2 n}{2} = n.

So it works!

Tn2−Tn−12=(Tn−Tn−1)(Tn+Tn−1)=n⋅n2=n3. \begin{aligned} T_n^2 - T_{n-1}^2 &=& (T_n - T_{n-1})(T_n + T_{n-1}) \\ &=& n \cdot n^2 \\ &=& n^3. \end{aligned}

Voila!