# Find the nth term

1. Apr 17, 2005

### misogynisticfeminist

I've got a very tricky question on my hands.

A set of integers are grouped as follows

$$(1), (2,3,4), (5,6,7,8,9),...,$$ until the nth bracket.

I have found the total integers in the first (n-1) brackets and it is $$(n-1)^2$$. The next part of the question is to show that the first number in the first term in the nth bracket is $$n^2-2n+2$$. What i did was to first write out the sequence representing the first term in each bracket,

$$1,2,5,10,17,...$$

but i can't seem to find any pattern with this sequence but have only seen that their difference is an arithmetic progression. How do I go about this question?

2. Apr 17, 2005

### quasar987

You hardly have to show this. There are (n-1)² terms contained in the first (n-1) brackets. So the LAST term of the (n-1)th bracket is the number (n-1)². This makes the first term of the nth bracket (n-1)² + 1 = n² - 2n +2

3. Apr 17, 2005

### dextercioby

And that sequence of #-s $1,2,5,10,17,26,...$ can be described by

$$a_{n+1}=a_{n}+2n-1$$

Daniel.