I am having trouble finding the nth term of a sequence. I understand how to find the second difference but the rule with A confuses me.

Question

There are $6$ terms listed from term $1$ to term $6$ as:

$1,3,6,10,15,21$

from this came the first difference:

$2,3,4,5,6$

and the second difference is:

$1,1,1,1$

Does this change anything with the nth term equation?

Answer 1

You can list the terms as follows: $a_{0} = 0, a_{1} = 1, a_{2} = 3, a_{3} = 6, a_{4} = 10, a_{5} = 15, a_{6} = 21, \cdots.$

You have correctly noted that $a_{n} - a_{n -1} = n$.

You can rewrite that as $\boxed{a_{n} = n + a_{n-1}}$.

Now, in the above recurrence formula, you can replace $a_{n-1}$ with $n -1 + a_{n-2}$ to get $\boxed{a_{n} = n + (n -1) + a_{n-2}}$. If you keep doing this, you will get $\boxed{a_{n} = \sum_{k =1}^{n} k + a_{0} = \frac{n (n +1)}{2}}$.

Answer 2

If you start the sequence instead at $n=0$, the resulting difference table is: \begin{matrix} \color{red}{0} &1 &3 &6 &10 &15 &21 \\ \color{blue}{1} &2 &3 &4 &5 &6 \\ \color{orange}{1} &1 &1 &1 &1 \end{matrix} You can read off the formula from the leftmost column, yielding $$\color{red}{0}\binom{n}{0} + \color{blue}{1}\binom{n}{1} + \color{orange}{1}\binom{n}{2} = 0 + n + \frac{n(n-1)}{2} = \frac{n(n+1)}{2}$$