Quadratic sequence nth terms

(a) $n^2 + 2n$

1. Differences: 5, 7, 9, 11. Second difference: 2, so the $n^2$ coefficient is $2 \div 2 = 1$.
   $$n^2: \ 1, 4, 9, 16, 25$$
2. Subtract $n^2$ from each term.
   $$3 - 1 = 2, \ 8 - 4 = 4, \ 15 - 9 = 6, \ \ldots \Rightarrow 2n$$
3. So the $n$th term is $n^2 + 2n$.

• M1 Finding the second difference 2 and the term $n^2$.
• M1 Subtracting $n^2$ to get the linear part $2, 4, 6, \ldots$
• A1 The correct answer, $n^2 + 2n$.

(a) $2n^{2} + n + 3$

1. $$2n^{2}$$
2. $$n+3$$
3. The first differences are 7, 11 and 15, so the second difference is 4.
4. For $an^{2}$ + bn + c, the second difference is 2a; hence a = 2.
5. Subtract $2n^{2}$ from the terms to obtain 4, 5, 6 and 7, whose nth term is n + 3.
6. So the nth term is $2n^{2} + n + 3$.

• P1 Establishing $2n^{2}$ or an equivalent valid method.
• P1 Establishing $n+3$ or an equivalent valid method.
• A1 Correct answer: $2n^{2} + n + 3$

(a) $1$

1. Complete the square.
   $$n^2 - 6n + 10 = (n - 3)^2 + 1$$
2. $(n - 3)^2 \ge 0$, and it is 0 when $n = 3$. So the smallest term is the 3rd term, 1.

• P1 Writing $(n - 3)^2 + 1$, or listing terms $5, 2, 1, 2, 5$.
• A1 The correct answer, $1$.

(b) Each term is a square plus 1, so it is at least 1.

1. Every term equals $(n - 3)^2 + 1$. A square is never negative, so every term is at least 1.

• C1 Using $(n - 3)^2 + 1 \ge 1$ (a square is never negative).

(a) $2n^{2}+6$

1. The constant second difference gives twice the quadratic coefficient.
   $$2\times 2$$
2. Subtract the quadratic part and fit the remaining linear sequence.
   $$(2n^{2}+6)-2n^{2}$$
3. Therefore $2n^{2}+6$.

• M1 The constant second difference gives twice the quadratic coefficient.
• M1 Subtract the quadratic part and fit the remaining linear sequence.
• A1 Correct answer: $2n^{2}+6$

(b) $206$

1. Substitute n=10 into all terms of the rule.
   $$2\times 10^{2}+0\times 10+6$$
2. Therefore $206$.

• M1 Substitute n=10 into all terms of the rule.
• A1 Correct answer: $206$

(a) $3n^{2}-n+9$

1. Use the second difference to write the form $an^2 + bn + c$.
   $$2\times 3$$
2. Use the two known terms to solve for the linear coefficient and constant.
   $$2b+c=7$$
3. Therefore $3n^{2}-n+9$.

• M1 Use the second difference to write the form $an^2 + bn + c$.
• M1 Use the two known terms to solve for the linear coefficient and constant.
• A1 Correct answer: $3n^{2}-n+9$

(b) $21$

1. The 20th term equals the threshold and all later differences are positive.
   $$3\times 20^{2}-20+9=1189$$
2. Therefore $21$.

• M1 The 20th term equals the threshold and all later differences are positive.
• A1 Correct answer: $21$

(a) $64$

1. Substitute n = 5, squaring before multiplying by 2.
   $$2\times 5^{2}+3\times 5-1$$
2. Therefore $64$.

• M1 Substitute n = 5, squaring before multiplying by 2.
• A1 Correct answer: $64$

(b) $10$

1. Form and solve a quadratic equation in n.
   $$2n^{2}+3n-230=0$$
2. Therefore $10$.

• M1 Form and solve a quadratic equation in n.
• A1 Correct answer: $10$

(a) $a = 2, b = 1, c = 1$

1. Differences 7 and 11; second difference 4, so $2a = 4$.
   $$a = 2$$
2. First difference: the gap from term 1 to term 2 is $3a + b$.
   $$3(2) + b = 7 \Rightarrow b = 1$$
3. First term:
   $$a + b + c = 4 \Rightarrow c = 1$$

• P1 Finding $a = 2$ from the second difference.
• P1 Finding $b = 1$ (from $3a + b = 7$ or by subtracting $2n^2$).
• A1 The correct answer, $(2, 1, 1)$.

(b) The 14th term

1. Solve.
   $$2n^2 + n + 1 = 407 \Rightarrow 2n^2 + n - 406 = 0$$
2. Factorise.
   $$(2n + 29)(n - 14) = 0 \Rightarrow n = 14$$
3. Check: $2 \times 196 + 14 + 1 = 407$.

• P1 Forming $2n^2 + n - 406 = 0$.
• A1 The correct answer, $n = 14$.

(a) $30$

1. $$2b+c=5$$
2. $$5b+c=17$$
3. $$29^{2}+4\times 29-3$$
4. $$30^{2}+4\times 30-3$$
5. A constant second difference of 2 means the nth term is $n^{2} + bn + c$.
6. The given terms give 2b + c = 5 and 5b + c = 17. Subtract to get b = 4, then c = $-3.$
7. The rule is $n^{2}$ + 4n $-$ 3. Its consecutive increases are 2n + 5, which are positive for n $\ge$ 1.
8. The 29th term is $29^{2}$ + 4 $\times$ 29 $-$ 3 = 954; the 30th is $30^{2}$ + 4 $\times$ 30 $-$ 3 = 1017.
9. Because the terms increase, the first term above 1000 is the 30th.

• P1 Establishing $2b+c=5$ or an equivalent valid method.
• P1 Establishing $5b+c=17$ or an equivalent valid method.
• P1 Establishing $29^{2}+4\times 29-3$ or an equivalent valid method.
• P1 Establishing $30^{2}+4\times 30-3$ or an equivalent valid method.
• A1 Correct answer: $30$