Question

the series of natural numbers is divided into groups (1), (2,3,4) ,(5,6,7,8,9)..........show that sum of numbers in nth group is (n-1)^3 +n^3

Answer 1

also the sum of number of natural number in each group is equal to the last number of that group i.e.,
1 + 2 = 3 = last number of second group
1 = 1 = last number of first group
1 + 2 + 3 = 6 = last number of third group
⇒ 1 + 2 + ..... + 49 = last number of 49th group =
⇒ First number of 50th group = 1225 + 1 = 1226
⇒ 50th group contains  50 consecutive natural numbers i.e., with common difference 1 and first term 1226
Thus, sum of numbers in 50th group =
In the mentioned query it can be seen that the number of natural numbers in the first, second, third,....., nth groups are 1, 2, 3, ...., n