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: $$ \frac {n^2(n^2+1)} 2 - \frac {(n-1)^2((n-1)^2 + 1)} 2 $$

Simplify as you wish.

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  • $\begingroup$ What is the last number in the nth group? $\endgroup$ – Adienl Jun 14 '13 at 13:14
  • $\begingroup$ $n^2$ is the last number as can be seen easily from the sequence $\endgroup$ – Parth Thakkar Jun 14 '13 at 13:15
  • $\begingroup$ Yeah, I see it too. $\endgroup$ – Adienl Jun 14 '13 at 13:17

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