Question

On the ground are placed n stones, the distance between the first and second is one yard, between the 2nd and 3rd is 3yds, between the 3rd and 4th, 5yds, and so on. How far will a person have to travel who shall bring them one by one to a basket placed at the first stone?

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

Answer 1

Answer: (B)

$\frac{1}{3}(n-1)n(2n-1)$ yds.

Given ’n’ number of stones with (2k-1) distance between each successive stones,

Where $k=1,2,3,...........n$

Total distance covered to bring them one by one to the basket is

$S=0+2+8+18+\cdots +T_n$
Where$T_n$ is the distance covered f or the last stone to collect it
$T_n=2(1+3+5+\cdots +(2n-1\left)\right)$
$\Rightarrow T_n=2\times \left(\frac{n}{2}\right(2+2(n-1))=2n^2$
$\Rightarrow S=2(1+4+9+\cdots +n^2)$
$\Rightarrow S=2\left[\frac{n(n+1)(2n+1)}{6}\right]=\frac{n(n+1)(2n+1)}{3}$
Replace $n$ by $(n-1)$ because there are n-1 distances between n stones
$\Rightarrow S=\frac{(n-1)n(2n-1)}{3}$ yds