Question

The series of natural numbers is divided into groups $\left(1\right),(2,3,4),(3,4,5,6,7),(4,5,6,7,8,9,10),...$ Find the sum of the numbers in nth group.

• A. ${[n+1]}^2$
• B. ${[n-1]}^2$
• C. ${[2n-1]}^2$
• D. ${[2n+1]}^2$

Answer 1

Answer: (C)

${[2n-1]}^2$

$\left(1\right),(2,3,4),(3,4,5,6,7),(4,5,6,7,8,9,10),...$
nth group contains $2n-1$ terms which are in A.P. of common difference $1$ and
first term of successive groups are $1,2,\mathrm{3....}$ and hence of nth group is $n$.
Sum of terms of nth group $=\frac{2n-1}{2}[2.n+(2n-1-1).1]=\frac{2n-1}{2}[4n-2]={[2n-1]}^2$
Hence, option $C$.