Here is an interesting case of numerical explorations leading to mathematical results. Grade-eleven me was interested in the relationship between parabolas and exponentials, and thought that the infinite series \[S = \sum_{n=1}^\infty \frac{n^2}{2^n}\] had something to do with it. Aiming to sum this series, I took the sequence of partial sums Sn and "cleared denominators" to obtain the integer sequence Tn = 2nSn. Then I (did the only thing I knew how to do: I) made a difference table…

n Tn ΔTn Δ2Tn Δ3Tn
1 1
5
2 6 10
15 12
3 21 22
37 24
4 58 46
83 48
5 141 94
177 96
6 318 190
367
7 685
…stopping at Δ3 because I recognized 12, 24, 48, 96, as a geometric progression! Namely, 6·2, 6·22, 6·23, 6·24. Then I noticed that the last two columns differ by 2 = 12 − 10 = 24 − 22 = 48 − 46 = 96 − 94. The next set of slanted differences were 10 − 5 = 5, 22 − 15 = 7, 46 − 37 = 9, 94 − 83 = 11, 190 − 177 = 13: odd numbers! Continuing leftward, 5 − 1 = 4, 15 − 6 = 9, 37 − 21 = 16, 83 − 58 = 25, … squares!

Now, this was all just detective-work: the fact that Tn appeared to be a linear combination of a square, an odd number, two, and six times a power of two had only been shown for n = 1, …, 4. Nevertheless, hopping back along the columns from Δ3Tn led me to conjecture that

Tn = 6·2n − (n + 1)2 − (2n + 3) − 2,

because, for example, T3 = 21 = 48 − 2 − 9 − 16 = 6·23 − (2·3 + 3) − (3 + 1)2. The formula for Tn simplified to 6·2nn2 − 4n − 6, yielding

Sn = 6 − (n2 + 4n + 6) / 2n

This final formula is correct: it can be verified by induction (as it was by grade-eleven me). (To do this, call the right-hand side Sn′, show that S1 = S1′, and that Sn+1′ = Sn′ + (n + 1)2 / 2n.) Anyway, taking the limit as n → ∞ reveals that S = 6. Six!