Discussion
A person with list of n items, collects 1st item one (unit) time, 2nd item two (2 units) times, 3rd item three (3 units) times, ........ rth items r times (r units) ....... to his cart while going around the mart ......... number of units will be in his cart by the time he reaches to billing counter, if he has list of 6(n = 6) items for that day procurement.
Answer: C
The cart holds 21 units.
The person takes 1 unit of the first item, 2 of the second, and so on up to r units of the rth, so the total for n items is the sum of the first n natural numbers:
\[ \begin{aligned} 1 + 2 + \dots + n &= \frac{n(n+1)}{2} \\ &= \frac{6 \times 7}{2} = 21 \end{aligned} \]
6 counts only the items, not the units. 42 is \(n(n+1)\), forgetting to halve, and 14 matches no step of the sum.
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