Question

On the ground $n$ stones are placed. The distance between the first and the second is one metre, between second and third is three metre, between third and fourth is five metres and so on. How far will a person have to travel to bring them one by one to the basket placed at the first stone if he starts to from the basket.

• A. $\frac{(n-1)n(2n-1)}{6}$
• B. $\frac{(n-1)(2n-1)}{6}$
• C. $\frac{(n-1)(2n-1)}{3}$
• D. $\frac{(n-1)n(2n-1)}{3}$

Answer 1

The correct option is D $\frac{(n-1)n(2n-1)}{3}$

Distance covered by the person to bring first stone to the basket $D_1=0$
Distance covered by the person to bring second stone to the basket
$D_2=A_1{A}_2+A_2{A}_1=1+1=2$ metres
Distance covered by the person to bring third stone to the basket
$D_3=A_1{A}_3+A_3{A}_1=2(1+3)$ metres
Distance covered by the person to bring fourth stone to the basket
$D_4=A_1{A}_4+A_4{A}_1=1+1=2(1+3+5)$ metres
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Hence total distance covered by the person to bring $n$ stones to the basket one by one,
$D=D_1+D_2+D_3+D_4+....n$ terms
$=0+2+2(1+3)+2(1+3+5)+...(n-1)$ terms
$=2[1+(1+3)+(1+3+5)+......+(n-1\left)terms\right]$
$n$th term of the series,
$T_n=2(1+3+5+......nterms)$
$=2\frac{n}{2}\left\{\right(2.1+(n-1)2\left)\right\}$
$T_n=2n^2$
$\therefore D=$Sum of $(n-1)$ terms $={\sum }_{n=1}^{n-1}2n^2$
$=2{\sum }_{n=1}^{n-1}{n}^2=2.\frac{(n-1)n(2n-1)}{6}$
$=\frac{(n-1)n(2n-1)}{3}$