If the set of natural numbers is partitioned | vedclass

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(A) The $n^{th}$ set $S_n$ contains $n$ terms.The first term of $S_n$ is given by $a_n = 1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.For $n = 50$,t

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$62525$

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(A) The $n^{th}$ set $S_n$ contains $n$ terms.The first term of $S_n$ is given by $a_n = 1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.For $n = 50$,the first term is $a_{50} = 1 + \frac{49 	imes 50}{2} = 1 + 1225 = 1226$.The set $S_{50}$ contains $50$ consecutive integers starting from $1226$,so $S_{50} = \{1226, 1227, \dots, 1275\}$.The sum of these $50$ terms is an arithmetic progression sum: $Sum = \frac{n}{2}(2a + (n-1)d)$.$Sum = \frac{50}{2}(2 	imes 1226 + (50-1) 	imes 1) = 25(2452 + 49) = 25(2501) = 62525$.

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

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