The series of natural numbers is divided into groups as follows ; (1), (2,3),(4,5,6), (7,8,9,10) and so on. The sum of the numbers in the Nth group is
In the mentioned query it can be seen that the number of natural numbers in the first, second, third,....., nthgroups are 1, 2, 3, ...., n
also the sum of number of natural number in each group is equal to the last number of that group i.e.,
1 = 1 = last number of first group
1 + 2 = 3 = last number of second group
1 + 2 + 3 = 6 = last number of third group
⇒ 1 + 2 + ..... + n-1= last number of (n-1)th group = (n-1)(n-1+1)/2 =
n(n-1)/2
Therefore first number of nth group= n(n-1)/2 + 1
Nth group contains N consecutive natural numbers i.e., with common difference 1 and first term=n(n-1)/2 +1
Thus the sum of numbers in the nth group is =
n/2*[ (n[n-1]/2 + 1) + (n-1)*1]
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