TRIANGULAR NUMBERS

Find the 50th triangular number. The first four triangular numbers are T₁=1, T₂=3, T₃=6, and T₄=10.

T₁ = 1 = 1

T₂ = 1 + 2 = 3

T₃ = 1 + 2 + 3 = 6

T₄ = 1 + 2 + 3 + 4 = 10

T₅ = 1 + 2 + 3 + 4 + 5 = 15

T_n = 1 + 2 + ... + (n-2) + (n-1) + n = ?

Realizing that T_n is defined by the sum of natural numbers from 1 to n was the primary step in solving this problem. It did not take me too long to see this, so I can't really find a more intuitive solution step prior to this one. This pattern lends itself very easily to a recursive formula, since each new triangular number is merely the previous triangular number plus the new term, or in a simpler symbolic expression: T_n = T_n-1 + n.

Unfortunately, this pattern and idea is useless to us for finding the 50th triangular number unless we know the 49th triangular number. At the most primitive level, we can solve this problem by summing the numbers from 1 to 50 (on calculator).= 1275. This was VERY annoying to do. It took me multiple times to get the same sum twice (i.e. to verify that I had done it correctly). At this point, the first thing that comes to mind is one of the methods of summation learned in the Lion problem. Specifically, add the terms successively from opposite ends of the sequence so that you get the same sum with each addition. In the Lion problem, the sum always equaled n-1, BUT the summation went from 1 to n-2. This time, the summation goes from 1 to n, and as we will discover, the sum will equal n+1 for each pair.

The only remaining question, then, is exactly how many times do we count the number 51? Here we must consider the two cases where n is an odd or even number.

If n is odd:

1 + 2 + . . . + (-1) + + (+1) + (+2) . . . + (n-1) + n

Above you can see that you keep adding pairs until you get to the n/2 term. In other words, when n is odd, you add n+1, n/2 times. Or: = (n+1)(n/2).

If n is even, the sequence looks slightly different:

1 + 2 + . . . + ( -1) + + + ( +1) + ( +2) + . . . + (n-1) + n

Here, observe that we can only sum pairs to the (n-1)/2 term, and then there is an extra term, (n+1)/2 to be added to the sum. So: = (n+1)[(n-1)/2] + (n+1)/2.

Now, the only question is: Does (n+1)(n/2) = (n+1)[(n-1)/2] + (n+1)/2 ?

Observe,

(n+1)[(n-1)/2] + (n+1)/2 =

[(n+1)(n-1)]/2 + (n+1)/2 =

[(n+1)(n-1) + (n+1)]/2 =

[(n+1)(n-1+1)]/2 =

[(n+1)n]/2 =

(n+1)(n/2).

So, our closed formula is, in all cases, T_n = (n+1)(n/2)

Someone else in the class came up with a very interesting alternative way to find (visualize) the recursive formula. Consider,

If T_n is visualized as a figure n², then T_n = n² - T_n-1 . The interesting question here is: How does n² - T_n-1 = T_n-1 + n ? Observe:

T_n = n² - T_n-1 T_n = T_n-1 + n

T_n + T_n-1 - n² = 0 T_n-1 + n -T_n = 0

T_n +T_n-1 - n² = T_n-1 + n - T_n

2T_n - n² -n = 0

2T_n = n² + n

2T_n = n(n+1)

T_n = n(n+1)/2.

SUMMARY OF SOLUTIONS:

Numerical Solution

T₅₀ = 1275

Recursive Formula

T_n = T_n-1 + n

Closed Formula

T_n = (n+1)n/2