This is recursive because here you have terms introduce to "previous" terms, till we reach the base situation $a_0$. Your definition of $a_n = 2n+1$ is no recursive because you"re no recursing come a base case; your definition is just a simple computation.

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**b)** $a_0 = 1$; $a_n+1 = a_n \cdot 3$

This provides us $1, 3, 9, \ldots$ so it works. (Noting the $3^0=1$.)

An concern that the welcomed answer doesn"t address is that the concern asks because that a set, no a sequence.

The appropriate answer is:

**Base step**:

$3 \in S$

**Recursive step**:

$x \in S \implies 3x \in S$

This reads that 3 is in collection $S$. If x is in $S$, climate $3x$ is additionally in $S$, for this reason recursively defining this set because it builds $\\space3,\space9\space(3(3)),\space27\space(3(9)),\space81\space(3(27))\space...\space\$

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