If we call the terms a₀, a₁, a₂, a₃, a₄, a₅ and group them in pairs: (a₀, a₁), (a₂, a₃), (a₄, a₅) we can relate the second member of each pair to its partner: a[2k+1]=a[2k](2k+1) where the square brackets represent the subscript and k starts at 0. So, for example, if k=1, a₃=3a₂ and if k=2, a₅=5a₄.

Also a[2k]=2^(2^k), so, for example, if k=0, a₀=2^(2⁰)=2¹=2 and if k=1, a₂=2^(2¹)=2²=4, and a₁=a₀×1=2; if k=1, a₃=3a₂=3×4=12. So we have the series: 2, 2, 4, 12, 16, 80, the given series.

Therefore, a₆ is found from k=3: 2^(2³)=2⁸=256. The series becomes 2, 2, 4, 12, 16, 80, 256. (The next term would be 7×256=1792.)

This logic can be simplfied:

First, write down the first, third and fifth term but leave a gap between them:

2 | 4 | 16 |

(2 multiplied by itself is 4, 4 multiplied by itself is 16.)

Next, write the numbers 1 to 6 underneath, including the gaps:

2 | 4 | 16 | |||

1 | 2 | 3 | 4 | 5 | 6 |

To fill in the gaps, multiply the number on the top line by the one beneath it and put the result where the gap next to it is:

2 | 2 | 4 | 12 | 16 | 80 |

× | =↑ | × | =↑ | × | =↑ |

1 | 2 | 3 | 4 | 5 | 6 |

So, 2×1=2 goes into the first gap; 4×3=12 goes into the next gap; 16×5=80. That gives us the series. So to continue, we would, by this logic, need 16 multiplied by itself=256.

There is no teacher to ask Im studying for college Acceptance exam And that question gets stuck in my head Anyway thanks for your time and help

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