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Sums of Sums of Triangular Numbers (Posted on 2020-09-05)

Let T₁(n) be the n-th Triangular number, the sum of 1 + 2 + ... + n.

T₂(n) is the sum of the first n Triangular numbers.

T₃(n) is the sum of the first n T₂(n) numbers.

...

T_k(n) is the sum of the first n T_k-1(n) numbers.

Please derive closed form expressions for each of T₁(n), T₂(n), and T₃(n). Describe the technique you used for the derivation.

Bonus: what is the formula for T_k(n)?

Bonus 2: prove the validity of your formula for T_k(n).

See The Solution Submitted by Larry

Rating: 4.5000 (2 votes)

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Solution

Comment 2 of 2 |

Re-indexing k to (k-1) for convenience later:

1, 1, 1, 1, 1, 1, T0(n) = 1

1, 2, 3, 4, 5, 6, T1(n) = n/1

0, 1, 3, 6, 10, 15, T2(n) = n*(n-1)/2

0, 0, 1, 4, 10, 20, T3(n) = n*(n-1)*(n-2)/6

0, 0, 0, 1, 5, 15, 35, T4(n) = n*(n-1)*(n-2)*(n-3)/24

0, 0, 0, 0, 1, 6, 21, 56 T5(n) = n*(n-1)*(n-2)*(n-3)*(n-4)/120

So Tk(n) = (n!/(n-k)!)/k! or (n!)/(k!(n - k)!)

Technique: computation of small terms plus observation.

Since the formula for Tk(n) is simply a rewriting of the observed results in generalised form, the issue of validity doesn't really arise.

nb the vertical columns of the above table comprise Pascal's Triangle; unsurprisingly, since (n!)/(k!(n - k)!) = C(n,k)

Edited on November 1, 2020, 12:48 am

Posted by broll on 2020-11-01 00:21:07

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