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.
21
The r-th item on the list is picked r times, so the total number of units is the sum of the first n natural numbers, \(\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}\).
Step 1 — For the six items: 1 + 2 + 3 + 4 + 5 + 6 units.
Step 2 — Using the formula, \(\dfrac{6 \times 7}{2} = 21\) units.
Six is only the number of items on the list, fourteen does not correspond to any partial sum of the series, and forty-two is \(6 \times 7\) without dividing by 2, so it double-counts every unit.
21 option is correct.
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