JAMB 2021 Mathematics · Question 8

What is the n-th term of the sequence 2, 6, 12, 20...?

Options

  1. A.4n - 2
  2. B.2(3n - 1)
  3. C.n2 + n
  4. D.n2 + 3n + 2

Answer: C. n2 + n

Explanation written with AI and checked against the answer key

Analysis

To find the nn-th term of the sequence 2,6,12,20,…2, 6, 12, 20, \dots, we need to identify the pattern or the rule that generates each term based on its position (nn).

Concept

A sequence can often be solved by looking at the relationship between the position nn (where n=1,2,3,4,…n = 1, 2, 3, 4, \dots) and the value of the term.

  • For n=1n = 1, the term is 22.
  • For n=2n = 2, the term is 66.
  • For n=3n = 3, the term is 1212.
  • For n=4n = 4, the term is 2020.
Step-by-Step Solution

Step 1: Test the options by substituting n=1,2,3n = 1, 2, 3

  • Option A (4n−24n - 2): • n=1:4(1)−2=2n=1: 4(1) - 2 = 2 (Correct) • n=2:4(2)−2=6n=2: 4(2) - 2 = 6 (Correct) • n=3:4(3)−2=10n=3: 4(3) - 2 = 10 (Incorrect, the sequence has 12)
  • Option B (2(3n−1)2(3n - 1)): • n=1:2(3(1)−1)=2(2)=4n=1: 2(3(1) - 1) = 2(2) = 4 (Incorrect, the sequence starts with 2)
  • Option C (n2+nn^2 + n): • n=1:(1)2+1=1+1=2n=1: (1)^2 + 1 = 1 + 1 = 2 (Correct) • n=2:(2)2+2=4+2=6n=2: (2)^2 + 2 = 4 + 2 = 6 (Correct) • n=3:(3)2+3=9+3=12n=3: (3)^2 + 3 = 9 + 3 = 12 (Correct) • n=4:(4)2+4=16+4=20n=4: (4)^2 + 4 = 16 + 4 = 20 (Correct)
  • Option D (n2+3n+2n^2 + 3n + 2): • n=1:(1)2+3(1)+2=1+3+2=6n=1: (1)^2 + 3(1) + 2 = 1 + 3 + 2 = 6 (Incorrect, the sequence starts with 2)

Step 2: Verification The formula n2+nn^2 + n (which can also be written as n(n+1)n(n+1)) perfectly generates every term in the given sequence.

Conclusion

The formula that satisfies the sequence 2,6,12,202, 6, 12, 20 is n2+nn^2 + n.

Exam Tip

When you see a sequence where the differences between terms are increasing (in this case: 6−2=46-2=4, 12−6=612-6=6, 20−12=820-12=8), it is usually a quadratic sequence. If the differences are 4,6,8…4, 6, 8 \dots, the second difference is constant (22), which confirms the formula will involve an n2n^2 term. Always test your options starting with n=1n=1 to save time!

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