Natural numbers have been divided into groups in the following way: `91,3),(5,7,9,11),(13,15,….,23),……..` Show that the sum of numbers in the nth group is `4n^3`

The sequence of odd natural numbers is divided into groups 1,3,5,7,9,11,"…." and so on. Show that the sum of the numbers in nth group is n^(3) .

Natural numbers are written as 1, (2,3), (4,5,6).. Show that the sum of number in the nth group is (n)/(2) (n +1) .

The series of natural numbers is divided into groups :(1);(2,3,4);(5,6,7,8,9)... and so on. The sum of numbers in the n^(th) group is

The series of natural numbers is divided into groups :(1);(2,3,4);(5,6,7,8,9)... and so on. The sum of numbers in the n^(th) group is

The cubes of the natural numbers are grouped as 1^(3),(2^(3),3^(3)),(4^(3),5^(3),6^(3)), . . . . . . . ., the the sum of the number in the n^(th) group, is

The natural numbers arearranged innthe form given below The rth group containing 2^(r-1) numbers. Prove that sum of the numbers in the nth group is 2^(n-2)[2^(n)+2^(n-1)-1] .

The series of natural numbers is divided into grounds as (1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10)………. If sum of elements in the 20^(th) ground is l , then l is equal to

Fourteen numbered balls (1, 2, 3, …, 14) are divided in 3 groups randomly. Find the probability that the sum of the numbers on the balls, in each group, is odd.

The number of different ways in which the first twelve natural numbers can be divided into three groups, each having four elements, such that the numbers in each group are in arithmetic progression is

The sequence N of natural numbers is divided into classes as follows. Show that the sum of the numbers in nth row is (2n^2+1)