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.

K but the answer in the book was after 80 comes 86

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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