nth Term of Quadratic Sequences [H]

nth Term of Quadratic Sequences

When a sequence has a constant second difference, its nth term formula is quadratic: \( T_n = an^2 + bn + c \). Deducing this formula from the sequence requires a systematic three-step method. The constant second difference is always equal to \( 2a \), which gives the value of \(a\) immediately.

Method

  1. Find \(a\): compute the constant second difference \(delta\). Then \( a = \delta/2 \).
  2. Subtract \(an^2\): write out the sequence \(an^2\) for \(n=1,2,3,ldots\) and subtract it from the original sequence term by term. The result is a linear sequence.
  3. Find the linear nth term: apply the standard linear method to the residual sequence to find \(bn+c\). Combine: \( T_n = an^2 + bn + c \).
n 1 2 3 4 5 Sequence 3 7 13 21 31 1st diffs 4 6 8 10 2nd diffs 2 2 2 ⇒ a=1 n² (a=1) 1 4 9 16 25 Residuals 2 3 4 5 6 Linear nth term: n+1 Tₙ = n² + n + 1

Worked Examples

Find the \(n\)th term of 5, 14, 27, 44, 65, …

First differences: 9, 13, 17, 21. Second differences: 4, 4, 4. So \(a=2\).

Subtract \(2n^2\): \(2,8,18,32,50\). Residuals: \(5-2=3, 14-8=6, 27-18=9, 44-32=12, 65-50=15\).

Residual sequence 3, 6, 9, 12, 15 has \(d=3\), so linear nth term = \(3n\).

\[ T_n = 2n^2 + 3n \]

Check: \(T_3=18+9=27 ✓\)

Find the \(n\)th term of 1, 4, 8.5, 14, 20.5, …

First differences: 3, 4.5, 5.5, 6.5. Second differences: 1, 1, 1. So \(a=\frac{1}{2}\).

Subtract \(\frac{1}{2}n^2\): \(\frac{1}{2},2,\frac{9}{2},8,\frac{25}{2}\). Residuals: \(1-0.5=0.5, 4-2=2, 8.5-4.5=4, 14-8=6, 20.5-12.5=8\). Hmm — let me recheck. \(\frac{n^2}{2}\) for \(n=1,2,3,4,5\): 0.5, 2, 4.5, 8, 12.5. Residuals: 0.5, 2, 4, 6, 8. Linear sequence with \(d=2\), first term 0.5: \(T_n=2n-1.5=\frac{4n-3}{2}\). Actually \(d = 1.5\). Residual: 0.5, 2, 3.5, 5, 6.5 — \(d=1.5\). Nth term of residuals: \(1.5n-1=\frac{3n-2}{2}\). So \(T_n = \frac{n^2}{2} + \frac{3n-2}{2} = \frac{n^2+3n-2}{2}\). Check \(T_2=\frac{4+6-2}{2}=4 ✓\).

For \(T_n=n^2+n+1\): find the 10th term. Is 157 in the sequence?
\[ T_{10}=100+10+1=111 \]

Is 157 in the sequence? \(n^2+n+1=157 Rightarrow n^2+n-156=0 Rightarrow (n+13)(n-12)=0 Rightarrow n=12 \text{ (positive root)}\). Integer ✓ — 157 is the 12th term.

 Key Takeaways

  • Constant second difference \(delta\) → quadratic sequence with \(a=delta/2\).
  • Method: find \(a\) → subtract \(an^2\) from each term → find linear nth term of residuals → combine.
  • The general form is \(T_n=an^2+bn+c\). The coefficient \(a\) is always half the constant second difference.
  • Always verify by substituting a term (ideally not one used in the derivation).
  • To check membership: set \(T_n=k\) and solve the resulting quadratic — a positive integer solution confirms membership.