Question

The series of natural numbers is written as follows:
1
234
5 6 7 8 9
.find the sum of the numbers in nth row​

Answer 1

(1),(2,3,4),(3,4,5,6,7),(4,5,6,7,8,9,10),...

nth group contains 2n−1 terms which are in A.P. of common difference 1 and

first term of successive groups are 1,2,3.... and hence of nth group is n.

Sum of terms of nth group =

$\frac{2n - 1}{2} [2.n+(2n−1−1).1]= \\ \\ </p><p></p><p> \frac{2n - 1}{2} [4n−2]=[2n−1]2</p><p></p><p>$