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 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) .

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)

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

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 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

Sum to n terms the series whose nth terms is (1)/(4n^(2)-1)

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] .

Two consecutive number p,p + 1 are removed from natural numbers. 1,2,3, 4……….., n-1, n such that arithmetic mean of these number reduces by 1, then (n-p) is equal to.

The total number of ways of selecting two numbers from the set {1,2, 3, 4, ........3n} so that their sum is divisible by 3 is equal to a. (2n^2-n)/2 b. (3n^2-n)/2 c. 2n^2-n d. 3n^2-n