Quadratic Sequences (n-th Term)

\( u_n=an^2+bn+c,\quad a=\tfrac{1}{2}\,\Delta^2,\quad b=\Delta u_1-3a,\quad c=u_1-a-b \)
Algebra GCSE
Question 9 of 20
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Find nth term for 4,7,12,19.

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( a=1,b=0,c=3. \)

Explanation

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Statement

A quadratic sequence has general term:

\[ u_n = an^2 + bn + c \]

where:

  • \(a=\tfrac{1}{2}\Delta^2\) (half the second difference),
  • \(b=\Delta u_1 - 3a\),
  • \(c=u_1-a-b\).

Why it’s true

  • Quadratic sequences have a constant second difference.
  • General form is quadratic: \(u_n=an^2+bn+c\).
  • Using the first few terms and differences, we solve for \(a,b,c\).

Recipe (how to use it)

  1. Write down first 3–4 terms of sequence.
  2. Find first and second differences.
  3. Calculate \(a=\tfrac{1}{2}\Delta^2\).
  4. Use \(b=\Delta u_1 - 3a\).
  5. Find \(c=u_1-a-b\).
  6. Write nth term as \(u_n=an^2+bn+c\).

Spotting it

If a sequence has a constant second difference, it’s quadratic.

Common pairings

  • Finding specific terms (e.g., 10th term).
  • Checking if a number is in the sequence.
  • Working backwards to find term positions.

Mini examples

  1. Sequence: 2, 5, 10, 17, 26.
    Second difference=2. \(a=1\). First difference=3. \(b=3-3(1)=0\). \(c=2-1-0=1\).
    nth term=\(n^2+1\).
  2. Sequence: 1, 6, 15, 28.
    Second difference=4. \(a=2\). \(\Delta u_1=5\). \(b=5-6= -1\). \(c=1-2-(-1)=0\).
    nth term=\(2n^2-n\).

Pitfalls

  • Forgetting to halve the second difference for \(a\).
  • Miscomputing b or c by plugging wrong terms.

Exam strategy

  • Check constant second difference before starting.
  • Always test formula with first few terms.

Summary

Quadratic sequences follow \(u_n=an^2+bn+c\). Use second differences to find \(a\), then compute \(b\) and \(c\).

Worked examples

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  1. Find nth term for 2,5,10,17,26.
    1. \( 2nd diff=2 → a=1 \)
    2. \( Δu1=3 \)
    3. \( b=3-3=0 \)
    4. \( c=2-1=1 \)
    Answer: \( n^2+1 \)
  2. Find nth term for 1,6,15,28.
    1. \( 2nd diff=4 → a=2 \)
    2. \( Δu1=5 \)
    3. \( b=5-6=-1 \)
    4. \( c=1-2+1=0 \)
    Answer: \( 2n^2-n \)
  3. Find nth term for 3,8,15,24.
    1. \( 2nd diff=2 → a=1 \)
    2. \( Δu1=5 \)
    3. \( b=5-3=2 \)
    4. \( c=3-1-2=0 \)
    Answer: \( n^2+2n \)
  4. Find nth term for 0,3,8,15,24.
    1. \( 2nd diff=2 → a=1 \)
    2. \( Δu1=3 \)
    3. \( b=3-3=0 \)
    4. \( c=0-1=-1 \)
    Answer: \( n^2-1 \)
  5. Find nth term for 7,12,19,28.
    1. \( 2nd diff=2 → a=1 \)
    2. \( Δu1=5 \)
    3. \( b=5-3=2 \)
    4. \( c=7-1-2=4 \)
    Answer: \( n^2+2n+4 \)
  6. Find nth term for 2,9,20,35.
    1. \( 2nd diff=4 → a=2 \)
    2. \( Δu1=7 \)
    3. \( b=7-6=1 \)
    4. \( c=2-2-1=-1 \)
    Answer: \( 2n^2+n-1 \)
  7. Find nth term for 5,14,27,44.
    1. \( 2nd diff=6 → a=3 \)
    2. \( Δu1=9 \)
    3. \( b=9-9=0 \)
    4. \( c=5-3=2 \)
    Answer: \( 3n^2+2 \)
  8. Find nth term for 4,11,22,37.
    1. \( 2nd diff=4 → a=2 \)
    2. \( Δu1=7 \)
    3. \( b=7-6=1 \)
    4. \( c=4-2-1=1 \)
    Answer: \( 2n^2+n+1 \)
  9. Find nth term for 10,19,32,49.
    1. \( 2nd diff=4 → a=2 \)
    2. \( Δu1=9 \)
    3. \( b=9-6=3 \)
    4. \( c=10-2-3=5 \)
    Answer: \( 2n^2+3n+5 \)
  10. Find nth term for 6,15,28,45.
    1. \( 2nd diff=4 → a=2 \)
    2. \( Δu1=9 \)
    3. \( b=9-6=3 \)
    4. \( c=6-2-3=1 \)
    Answer: \( 2n^2+3n+1 \)