The series of natural numbers is divided into groups 1;  2,3,4;  5,6,7,8,9; ...... and so on. Then the sum of the numbers in the nth group is

Last term (n−1)th  group  is  (n−1)2 ∴ first term is nth group is (n−1)2+1  and last term of nth group is n2Number of terms in nth group =(2n−1)∴  Sum of the numbers in the nth group =(2n−1)2{(n−1)2+1+n2}                                                                         =(2n−1)(n2−n+1) =2n3-2n2+2n-n2+n-1                                                                           =2n3-3n2+3n-1                                                                          =n3+(n−1)2