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If $a_1 = 1$ and $a_{n+1} = 2a_{n + 5}, n = 1, 2, \dots ,$ then $a_{100}$ is equal to

  1. $(5 × 2^99 – 6)$
  2. $(5 × 2^99 + 6)$
  3. $(6 × 2^99 + 5)$
  4. $(6 × 2^99 – 5)$
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