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Lesson: Recursive Formula of a Sequence

In this lesson, we will learn how to find the recursive formula of a sequence.

Worksheet • 4 Questions

Q1:

Find a recursive formula for the sequence 4 8 6 , 1 6 2 , 5 4 , 1 8 , 6 , 2 , 2 3 .

  • A π‘Ž = ο€Ό 1 3  π‘Ž 𝑛 𝑛 βˆ’ 1
  • B π‘Ž = ο€Ό 1 3  𝑛 1 βˆ’ 𝑛
  • C π‘Ž = ο€Ό 1 2  𝑛 𝑛 βˆ’ 1
  • D π‘Ž = ο€Ό 1 2  𝑛 𝑛
  • E π‘Ž = ο€Ό 1 3  𝑛 𝑛 + 1

Q2:

Write a recursive formula for the following sequence that is valid for 𝑛 β‰₯ 2 :

  • A π‘Ž = π‘Ž + 1 𝑛 𝑛 βˆ’ 1
  • B π‘Ž = π‘Ž βˆ’ 1 𝑛 𝑛 + 1
  • C π‘Ž = π‘Ž 𝑛 𝑛 + 1
  • D π‘Ž = π‘Ž βˆ’ 1 𝑛 𝑛 βˆ’ 1
  • E π‘Ž = π‘Ž 𝑛 𝑛 βˆ’ 1

Q3:

Using the absolute value function, write a recursive formula for the following sequence that is valid for 𝑛 β‰₯ 2 :

  • A π‘Ž = ( βˆ’ 1 ) ( | π‘Ž | + 1 ) 𝑛 𝑛 𝑛 βˆ’ 1
  • B π‘Ž = ( βˆ’ 1 ) ( | π‘Ž | + 1 ) 𝑛 𝑛 𝑛
  • C π‘Ž = ( 1 ) ( | π‘Ž | + 1 ) 𝑛 𝑛 𝑛 + 1
  • D π‘Ž = ( 1 ) ( | π‘Ž | βˆ’ 1 ) 𝑛 𝑛 𝑛 + 1
  • E π‘Ž = ( βˆ’ 1 ) ( | π‘Ž | βˆ’ 1 ) 𝑛 βˆ’ 𝑛 𝑛 βˆ’ 1

Q4:

Find a recursive formula for the sequence π‘Ž = 5 + 1 𝑛 𝑛 .

  • A π‘Ž = 5 π‘Ž βˆ’ 4 𝑛 + 1 𝑛 , π‘Ž = 6 1
  • B π‘Ž = 5 π‘Ž βˆ’ 6 𝑛 + 1 𝑛 , π‘Ž = 6 1
  • C π‘Ž = 5 π‘Ž + 4 𝑛 + 1 𝑛 , π‘Ž = 6 1
  • D π‘Ž = 5 π‘Ž + 6 𝑛 + 1 𝑛 , π‘Ž = 6 1
  • E π‘Ž = βˆ’ 5 π‘Ž + 4 𝑛 + 1 𝑛 , π‘Ž = 6 1

Find a recursive formula for the sequence π‘Ž = 2 ( 5 ) + 1 𝑛 𝑛 .

  • A π‘Ž = 5 π‘Ž βˆ’ 4 𝑛 + 1 𝑛 , π‘Ž = 1 1 1
  • B π‘Ž = 5 π‘Ž + 6 𝑛 + 1 𝑛 , π‘Ž = 1 1 1
  • C π‘Ž = 5 π‘Ž + 4 𝑛 + 1 𝑛 , π‘Ž = 1 1 1
  • D π‘Ž = 5 π‘Ž βˆ’ 6 𝑛 + 1 𝑛 , π‘Ž = 1 1 1
  • E π‘Ž = βˆ’ 5 π‘Ž + 4 𝑛 + 1 𝑛 , π‘Ž = 1 1 1

Find a recursive formula for the sequence π‘Ž = 1 3 ( 5 ) + 1 𝑛 𝑛 .

  • A π‘Ž = 5 π‘Ž βˆ’ 4 𝑛 + 1 𝑛 , π‘Ž = 8 3 1
  • B π‘Ž = 5 π‘Ž + 6 𝑛 + 1 𝑛 , π‘Ž = 8 3 1
  • C π‘Ž = 5 π‘Ž + 4 𝑛 + 1 𝑛 , π‘Ž = 8 3 1
  • D π‘Ž = 5 π‘Ž βˆ’ 6 𝑛 + 1 𝑛 , π‘Ž = 8 3 1
  • E π‘Ž = βˆ’ 4 π‘Ž + 5 𝑛 + 1 𝑛 , π‘Ž = 8 3 1

Find a recursive formula for the sequence π‘Ž = 5 π‘Ž + 1 𝑛 + 1 𝑛 and π‘Ž = 3 7 6 3 .

  • A π‘Ž = 3 ( 5 ) + 1 𝑛 𝑛
  • B π‘Ž = 3 . 0 8 ( 5 ) + 4 𝑛 𝑛
  • C π‘Ž = 3 . 0 0 8 ( 5 ) + 1 𝑛 𝑛
  • D π‘Ž = 3 ( 5 ) + 4 𝑛 𝑛
  • E π‘Ž = 2 . 9 3 6 ( 5 ) + 1 𝑛 𝑛
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