How do you find the nth term of the sequence #1, 3, 6, 10, 15,...#?

Question

Samantha Adkison · Accepted Answer

#a_n = 1/2n(n+1)#

These are recognisable as triangular numbers, but let's use a general method for finding matching polynomial formulas...

Write down the initial sequence:

#color(red)(1), 3, 6, 10, 15#

Write down the sequence of differences between consecutive pairs of terms:

#color(magenta)(2), 3, 4, 5#

Write down the sequence of differences of those differences:

#color(blue)(1), 1, 1#

Having reached a constant sequence, we can write down a formula for the #n#th term using the initial term of each of these sequences as a coefficient:

#a_n = color(red)(1)/(0!) + color(magenta)(2)/(1!)(n-1) + color(blue)(1)/(2!)(n-1)(n-2)#

#color(white)(a_n) = 1+2n-2+1/2n^2-3/2n+1#

#color(white)(a_n) = 1/2n^2+1/2n#

#color(white)(a_n) = 1/2n(n+1)#

Paisley Johnson · Answer

undefined To find the nth term of the sequence 1, 3, 6, 10, 15,..., we observe that each term is obtained by adding consecutive integers starting from 1. Specifically, the nth term can be expressed as the sum of the first n natural numbers.

The sum of the first n natural numbers is given by the formula $ S_n = \frac{n(n+1)}{2} $. Therefore, the nth term of the sequence is equal to $ S_n $ minus the sum of the first n-1 natural numbers.

To express this algebraically, we first find the sum of the first n-1 natural numbers using the formula $ S_{n-1} = \frac{(n-1)(n)}{2} $. Then, we find the nth term by subtracting $ S_{n-1} $ from $ S_n $.

So, the nth term of the sequence is given by:
$$ a_n = S_n - S_{n-1} = \frac{n(n+1)}{2} - \frac{(n-1)(n)}{2} $$

Simplifying the expression, we get:
$$ a_n = \frac{n^2 + n - n^2 + n}{2} = \frac{2n}{2} = n $$

Hence, the nth term of the sequence 1, 3, 6, 10, 15,... is simply n.