The series of positive multiples of 3 is divided into sets: 3, 6,9,12, 15,18,21,24,27,.

Question

The series of positive multiples of $3$ is divided into sets: $\{3\}, \{6,9,12\}, \{15,18,21,24,27\},\ldots$. Then the sum of the elements in the $11$th set is equal to \_\_\_\_\_\_\_\_.

Answer: 6993

Accepted Answer

The $k$th set has $2k-1$ terms. Up to the 10th set, the total number of terms (as multiples of $3$ indices) is $1+3+5+\cdots+19=100$. The 11th set has $21$ terms, so it consists of $3\times101, 3\times102, \ldots, 3\times121$.

$$\text{Sum} = 3(101+102+\cdots+121) = 3\times\frac{21(101+121)}{2} = 3\times21\times111 = 6993.$$

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