Algebra - Quadratic sequences (nth term)

Find the $n^{th}$ term formula for the sequence: $6, 13, 22, 33, 46, \dots$

1. Find the first differences: $13-6=7$, $22-13=9$, $33-22=11$, $46-33=13$.
2. Find the second differences: $9-7=2$, $11-9=2$, $13-11=2$. The constant second difference is $2$.
3. Find $a$: $2a = 2 \implies a = 1$
4. Find $b$: $3a + b = 7 \implies 3(1) + b = 7 \implies b = 4$
5. Find $c$: $a + b + c = 6 \implies 1 + 4 + c = 6 \implies c = 1$
6. The $n^{th}$ term is $T_n = n^2 + 4n + 1$.

We identify the second difference to determine the quadratic coefficient $a$, then use the linear relationships derived from the first term and the first difference to solve for $b$ and $c$.

Determine the $n^{th}$ term of the sequence: $2, 9, 20, 35, 54, \dots$

1. First differences: $7, 11, 15, 19$.
2. Second differences: $4, 4, 4$. Constant difference is $4$.
3. $a = \frac{4}{2} = 2$.
4. Subtract $an^2$ ($2n^2$) from the sequence:
   • For $n=1: 2 - 2(1)^2 = 0$
   • For $n=2: 9 - 2(2)^2 = 1$
   • For $n=3: 20 - 2(3)^2 = 2$
   • For $n=4: 35 - 2(4)^2 = 3$
5. The remaining sequence is $0, 1, 2, 3, \dots$ which is a linear sequence with $n^{th}$ term $n - 1$.
6. Combine the parts: $T_n = 2n^2 + n - 1$.

This method involves removing the quadratic component $an^2$ and finding the $n^{th}$ term of the resulting linear sequence ($bn + c$).