Question

The series of natural numbers is divided into groups $\left(1\right),(2,3,4),(3,4,5,6,7),(4,5,6,7,8,9,10),...$ Let the sum of the numbers in $nth$ group be $={[kn-m]}^2$.FInd $k+m$ ?

Answer 1

The number of terms in nth group is $2n-1$ which are 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
$=\left[(2n-1)/2\right][2.n+(2n-1-1).1]=\left[(2n-1)/2\right][4n-2]={[2n-1]}^2$
$\Rightarrow k+n=3$