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

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

Write down the initial sequence:

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

Write down the sequence of differences of those differences:

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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