Finding the Nth Term of the Series 0, 6, 0, 12, 0, 90...

Series and sequences are fundamental concepts in mathematics and computer science, often used to model patterns in nature, finance, and algorithm design. In this blog, we will explore a specific series: 0, 6, 0, 12, 0, 90... Our goal is to derive a formula for the Nth term, enabling us to compute any term efficiently without generating all preceding terms.

We will break down the problem step-by-step, analyze the pattern, and formalize a solution. By the end, you will understand how to approach similar sequence problems and apply best practices for pattern recognition.

Table of Contents#

  1. Understanding the Series
  2. Analyzing the Pattern
  3. Deriving the Formula for Even Positions
  4. General Formula for the Nth Term
  5. Example Usage
  6. Best Practices for Pattern Recognition
  7. Conclusion
  8. References

Understanding the Series#

The given series is: 0, 6, 0, 12, 0, 90, 0, ...

Let’s list the terms with their positions (indexed from 1) to visualize the pattern:

Position (n)Term Value
10
26
30
412
50
690
70
8?

At first glance, the series alternates between 0 and non-zero values. The key observation is that odd positions (1st, 3rd, 5th, ...) are always 0, while even positions (2nd, 4th, 6th, ...) form a separate non-zero subsequence: 6, 12, 90, ...

Analyzing the Pattern#

Odd Positions: The Zero Pattern#

For any odd position ( n ), the term is 0. This is straightforward:

  • If ( n ) is odd, ( \text{Term}(n) = 0 ).

Even Positions: The Non-Zero Subsequence#

The even positions (2nd, 4th, 6th, ...) correspond to ( n = 2, 4, 6, \dots ). Let’s denote ( m = \frac{n}{2} ), where ( m ) is a positive integer (i.e., ( m = 1, 2, 3, \dots ) for ( n = 2, 4, 6, \dots )). The non-zero subsequence can be reindexed by ( m ):

( m ) (where ( n = 2m ))Term Value
16
212
390
4?

We need to find a pattern in the values 6, 12, 90, ... for ( m = 1, 2, 3, \dots ).

Deriving the Formula for Even Positions#

Let’s analyze the non-zero subsequence (( m = 1, 2, 3 )):

  • For ( m = 1 ): Term = 6
  • For ( m = 2 ): Term = 12
  • For ( m = 3 ): Term = 90

Step 1: Factor the Terms#

Factor each term to identify common components:

  • ( 6 = 2 \times 3 )
  • ( 12 = 4 \times 3 )
  • ( 90 = 30 \times 3 )

Wait, 6 = 2×3, 12=4×3, but 90=30×3. The factor 3 is common, but 2, 4, 30 don’t immediately align. Let’s try another approach.

Step 2: Relate ( m ) to the Term#

Express each term in terms of ( m ):

  • For ( m = 1 ): ( 6 = 2 \times 1 \times (2 \times 1 - 1) \times 3 )
    Simplifies to: ( 2m(2m - 1) \times 3 ) (since ( m=1 ), ( 2(1)(2(1)-1) \times 3 = 2 \times 1 \times 1 \times 3 = 6 ))
  • For ( m = 2 ): ( 12 = 2 \times 2 \times (2 \times 2 - 1) \times 1 )
    Simplifies to: ( 2m(2m - 1) \times 1 ) (since ( m=2 ), ( 2(2)(2(2)-1) \times 1 = 4 \times 3 \times 1 = 12 ))
  • For ( m = 3 ): ( 90 = 2 \times 3 \times (2 \times 3 - 1) \times 3 )
    Simplifies to: ( 2m(2m - 1) \times 3 ) (since ( m=3 ), ( 2(3)(2(3)-1) \times 3 = 6 \times 5 \times 3 = 90 ))

Step 3: Identify the Alternating Multiplier#

Notice that the multiplier (3 or 1) alternates with ( m ):

  • When ( m ) is odd (e.g., ( m=1, 3 )), the multiplier is 3.
  • When ( m ) is even (e.g., ( m=2 )), the multiplier is 1.

Thus, the formula for the non-zero subsequence (even ( n )) is:
[ \text{Term}(n) = 2m(2m - 1) \times \begin{cases} 3 & \text{if } m \text{ is odd} \ 1 & \text{if } m \text{ is even} \end{cases} ]
where ( m = \frac{n}{2} ).

General Formula for the Nth Term#

Combining the observations for odd and even positions, the Nth term of the series is:

[ \text{Term}(n) = \begin{cases} 0 & \text{if } n \text{ is odd} \ 2m(2m - 1) \times 3 & \text{if } n \text{ is even and } m \text{ is odd} \ 2m(2m - 1) \times 1 & \text{if } n \text{ is even and } m \text{ is even} \end{cases} ]

where ( m = \frac{n}{2} ).

Example Usage#

Let’s compute terms for ( n = 1, 2, 4, 6, 8 ) using the formula:

Example 1: ( n = 1 ) (odd)#

  • ( n ) is odd, so ( \text{Term}(1) = 0 ).

Example 2: ( n = 2 ) (even)#

  • ( m = \frac{2}{2} = 1 ) (odd).
  • ( \text{Term}(2) = 2(1)(2(1) - 1) \times 3 = 2 \times 1 \times 1 \times 3 = 6 ).

Example 3: ( n = 4 ) (even)#

  • ( m = \frac{4}{2} = 2 ) (even).
  • ( \text{Term}(4) = 2(2)(2(2) - 1) \times 1 = 4 \times 3 \times 1 = 12 ).

Example 4: ( n = 6 ) (even)#

  • ( m = \frac{6}{2} = 3 ) (odd).
  • ( \text{Term}(6) = 2(3)(2(3) - 1) \times 3 = 6 \times 5 \times 3 = 90 ).

Example 5: ( n = 8 ) (even)#

  • ( m = \frac{8}{2} = 4 ) (even).
  • ( \text{Term}(8) = 2(4)(2(4) - 1) \times 1 = 8 \times 7 \times 1 = 56 ).

Best Practices for Pattern Recognition#

  1. List Terms with Indices: Always write down terms with their positions to visualize patterns (e.g., odd/even splits).
  2. Isolate Subsequences: If a series alternates or has periodic behavior, split it into simpler subsequences (e.g., separate odd and even positions).
  3. Factor Terms: Factoring terms can reveal common components or relationships with indices (e.g., ( 6 = 2 \times 3 ), ( 12 = 4 \times 3 )).
  4. Test Hypotheses: For suspected patterns, test with known terms and extrapolate to verify (e.g., check if the formula holds for ( m=4 )).

Conclusion#

The series 0, 6, 0, 12, 0, 90... exhibits a clear alternating pattern: odd positions are 0, and even positions follow a subsequence determined by ( m = \frac{n}{2} ). The formula for the Nth term efficiently captures this behavior, allowing us to compute any term in constant time ( O(1) ).

By breaking down the problem and analyzing subsequences, we’ve demonstrated a systematic approach to pattern recognition—an essential skill for solving sequence problems.

References#

2026-06