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
∑ π √ ≈
Find nth term for 2,9,20,35.
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)
- Write down first 3–4 terms of sequence.
- Find first and second differences.
- Calculate \(a=\tfrac{1}{2}\Delta^2\).
- Use \(b=\Delta u_1 - 3a\).
- Find \(c=u_1-a-b\).
- 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
- 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=\(n^2+1\).
- 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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Find nth term for 2,5,10,17,26.
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\( 2nd diff=2 → a=1 \)
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\( Δu1=3 \)
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\( b=3-3=0 \)
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\( c=2-1=1 \)
Answer:
\( n^2+1 \)
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Find nth term for 1,6,15,28.
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\( 2nd diff=4 → a=2 \)
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\( Δu1=5 \)
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\( b=5-6=-1 \)
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\( c=1-2+1=0 \)
Answer:
\( 2n^2-n \)
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Find nth term for 3,8,15,24.
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\( 2nd diff=2 → a=1 \)
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\( Δu1=5 \)
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\( b=5-3=2 \)
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\( c=3-1-2=0 \)
Answer:
\( n^2+2n \)
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Find nth term for 0,3,8,15,24.
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\( 2nd diff=2 → a=1 \)
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\( Δu1=3 \)
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\( b=3-3=0 \)
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\( c=0-1=-1 \)
Answer:
\( n^2-1 \)
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Find nth term for 7,12,19,28.
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\( 2nd diff=2 → a=1 \)
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\( Δu1=5 \)
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\( b=5-3=2 \)
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\( c=7-1-2=4 \)
Answer:
\( n^2+2n+4 \)
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Find nth term for 2,9,20,35.
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\( 2nd diff=4 → a=2 \)
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\( Δu1=7 \)
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\( b=7-6=1 \)
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\( c=2-2-1=-1 \)
Answer:
\( 2n^2+n-1 \)
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Find nth term for 5,14,27,44.
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\( 2nd diff=6 → a=3 \)
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\( Δu1=9 \)
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\( b=9-9=0 \)
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\( c=5-3=2 \)
Answer:
\( 3n^2+2 \)
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Find nth term for 4,11,22,37.
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\( 2nd diff=4 → a=2 \)
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\( Δu1=7 \)
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\( b=7-6=1 \)
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\( c=4-2-1=1 \)
Answer:
\( 2n^2+n+1 \)
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Find nth term for 10,19,32,49.
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\( 2nd diff=4 → a=2 \)
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\( Δu1=9 \)
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\( b=9-6=3 \)
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\( c=10-2-3=5 \)
Answer:
\( 2n^2+3n+5 \)
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Find nth term for 6,15,28,45.
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\( 2nd diff=4 → a=2 \)
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\( Δu1=9 \)
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\( b=9-6=3 \)
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\( c=6-2-3=1 \)
Answer:
\( 2n^2+3n+1 \)