Question

The series of natural numbers is divided into groups as follows; (1), (2, 3), (4, 5, 6), (7, 8, 9, 10) and so on. Find the sum of the numbers in the $n^{th}$ group is.

• A. $\frac{1}{2}\left[n\right(n^2+1\left)\right]$
• B. $\frac{n(n^2+1)}{4}$
• C. $\frac{2n(n^2+1)}{3}$
• D. $\frac{n^2(n^2+1)}{2}$

Answer 1

The correct option is A $\frac{1}{2}\left[n\right(n^2+1\left)\right]$

First element in the $n^{th}$ group:

$S_n=1+2+4+7+......a_n$ ......eqn(1)

$S_n=1+2+4+......+a_{n-1}+a_n$ eqn(2)

Eqn(1) - eqn(2) gives,

$0=1+1+2+3+........(n-1)-a_n{a}_n=1+\frac{(n-1)}{2}\left[2\right(1)+(n-2\left)1\right]a_n=1+\frac{(n-1)n}{2}andd=1$

Number of terms in $n^{th}$ group is $n$

Sum $=\frac{n}{2}[2[1+\frac{(n-1)n}{2}]+(n-1\left)\right(1\left)\right]=\frac{n}{2}[2+(n-1)n+(n-1\left)\right]=\frac{n}{2}(n^2-1+2)=\frac{n(n^2+1)}{2}$