Question

Let the natural numbers be divided into groups as $\left(1\right),(2,3,4),(5,6,7,8,9),...$ and so on. Then the sum of the numbers in the $n^{th}$ group is

• A. $n^3+(n+1{)}^3$
• B. $n^2-(n-1{)}^2$
• C. $(2n-1{)}^2$
• D. $n^3+(n-1{)}^3$

Answer 1

The correct option is D $n^3+(n-1{)}^3$

Clearly $n^{th}$ group contains $(2n-1)$ numbers.
The last term in each group are $1^2,2^2,3^2,...$
So, the last term in $n^{th}$ group is $n^2$.
Also, the first term of each group is one more than the last term of its previous group.
$\therefore$ The first term of the $n^{th}$ group is $(n-1{)}^2+1$.
Hence, the sum of the numbers in the $n^{th}$ group is given by
$\frac{2n-1}{2}\left[\right(n-1{)}^2+1+n^2](\because S_n=\frac{n}{2}(a+l\left)\right)=(2n-1\left)\right(n^2-n+1)=(n-1{)}^3+n^3$