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Triangular Numbers Sequence



Definition
Recursive Form
Derivation of the Closed Form
Properties



Dots Arranged in Triangles

The triangular numbers count objects arranged in growing equilateral triangles: one in the first row, two in the second, three in the third. Each triangular number is a partial sum of the natural numbers, giving them a direct connection to arithmetic series. They appear throughout mathematics — in binomial coefficients, in the relationship between consecutive squares, and in counting problems.



Definition

The nn-th triangular number TnT_n is the sum of the first nn positive integers:

Tn=1+2+3+⋯+n=n(n+1)2T_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}


The name comes from a geometric construction: arrange dots in rows of increasing length to form an equilateral triangle. Row 11 has 11 dot, row 22 has 22, row 33 has 33, and the total count after nn rows is TnT_n.

The first triangular numbers are:

1,3,6,10,15,21,28,36,45,55,…1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots


The closed form n(n+1)2\frac{n(n+1)}{2} follows from the arithmetic series formula with a1=1a_1 = 1 and d=1d = 1. The Gauss pairing argument applies directly: Tn=n2(1+n)T_n = \frac{n}{2}(1 + n).

Recursive Form

Each triangular number is the previous one plus a new row of nn dots:

T1=1,Tn=Tn−1+nT_1 = 1, \quad T_n = T_{n-1} + n


This recursive definition makes the growth pattern visible — the increments 1,2,3,4,5,…1, 2, 3, 4, 5, \ldots are themselves the natural numbers. The triangular numbers grow faster than a linear sequence (where increments are constant) but slower than an exponential one.

The increments form an arithmetic sequence with common difference 11. The triangular numbers are therefore the partial sums of an arithmetic sequence — they accumulate linearly growing additions.

Derivation of the Closed Form

The formula Tn=n(n+1)2T_n = \frac{n(n+1)}{2} can be derived by the pairing method. Write the sum forwards and backwards:

S=1+2+3+⋯+nS = 1 + 2 + 3 + \cdots + n


S=n+(n−1)+(n−2)+⋯+1S = n + (n-1) + (n-2) + \cdots + 1


Adding term by term: each pair sums to n+1n + 1, and there are nn pairs, giving 2S=n(n+1)2S = n(n+1) and S=n(n+1)2S = \frac{n(n+1)}{2}.

An alternative perspective: Tn=(n+12)T_n = \binom{n+1}{2}, the number of ways to choose 22 items from a set of n+1n+1. The binomial coefficient (n+12)=(n+1)!2!(n−1)!=(n+1)n2\binom{n+1}{2} = \frac{(n+1)!}{2!(n-1)!} = \frac{(n+1)n}{2} yields the same formula. This connection places triangular numbers inside combinatorics — they count unordered pairs. For example, T4=10T_4 = 10 equals the number of handshakes among 55 people.

Properties

The parity of triangular numbers follows a repeating cycle. TnT_n is odd when n≡1n \equiv 1 or 2(mod4)2 \pmod{4}, and even when n≡0n \equiv 0 or 3(mod4)3 \pmod{4}. The pattern of parities is odd, odd, even, even, odd, odd, even, even, …\ldots.

The sum of two consecutive triangular numbers is a perfect square:

Tn+Tn−1=n(n+1)2+(n−1)n2=n2T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{(n-1)n}{2} = n^2


Geometrically, two triangles of consecutive sizes fit together to form a square. This identity connects the two simplest figurate number sequences.

A number mm is triangular if and only if 8m+18m + 1 is a perfect square. Solving m=n(n+1)2m = \frac{n(n+1)}{2} for nn gives n=−1+1+8m2n = \frac{-1 + \sqrt{1 + 8m}}{2}, which is a positive integer precisely when 8m+18m + 1 is a perfect square. This provides a direct test: m=15m = 15 gives 8(15)+1=121=1128(15) + 1 = 121 = 11^2, so 1515 is triangular (T5=15T_5 = 15).

The sum of the first nn triangular numbers produces the tetrahedral numbers:

∑k=1nTk=n(n+1)(n+2)6\sum_{k=1}^{n} T_k = \frac{n(n+1)(n+2)}{6}


This extends the figurate number idea from two dimensions (triangles) to three (tetrahedra).