# The sum of the terms in the nth group

Let the natural numbers be divided into the following groups: {1}$,${2,3,4}$,${5,6,7,8,9}$.....$ What is the sum of the terms in the $n$th group?

I know that the number of terms in nth group will be $2n-1$. But, I am not able to get a general pattern for the terms in the nth group? Will it be different for when $n$ is even and when it is odd?

Why don't you just find the sum of numbers from 1 to the last number in the $n$th group, and then from it subtract the sum of the numbers from 1 to the last number in the $(n-1)$ th group?
That is: $$\frac {n^2(n^2+1)} 2 - \frac {(n-1)^2((n-1)^2 + 1)} 2$$
• $n^2$ is the last number as can be seen easily from the sequence – Parth Thakkar Jun 14 '13 at 13:15