Question

The series of natural numbers is divided into groups $\left(1\right),(2,3,4),(3,4,5,6,7),(4,5,6,7,8,9,10)$,______. Find the sum of the numbers in nth group.

Answer 1

As in part (a) the number of terms in nth group is $2n-1$ which are in A.P. of common difference $1$ and first term of successive groups are $1,2,3$, and hence of nth group is n. Sum of terms of nth group

$=\left[\right(2n-1\left)/2\right][2.n+(2n-1-1\left).1\right]$

$=\left[\right(2n-1\left)/2\right](4n-2)=(2n-1{)}^2$.