nth Term of Quadratic Sequences

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 3,5,7. Second difference: 2.
2. The aligned table shows the terms and their constant second difference.

3. The coefficient of n^2 is half the second difference: 2\div2=1, so start with n^2.
4. Subtract n^2 (that is 1,4,9,16) from the terms: 5-1=4, 8-4=4, 13-9=4, 20-16=4.
5. The remainder is always 4, so the nth term is n^2+4.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 5,7,9. Second difference: 2, so the coefficient of n^2 is 2\div2=1.
2. Subtract n^2: 3-1=2, 8-4=4, 15-9=6, 24-16=8.
3. The remainders 2,4,6,8 have nth term 2n.
4. Add the two parts: n^2+2n.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 6,10,14. Second difference: 4, so the coefficient of n^2 is 4\div2=2.
2. The aligned table shows the terms and their constant second difference.

3. The values of 2n^2 are 2,8,18,32. Each term of the sequence is 1 less than these.
4. The nth term is 2n^2-1.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 2,4,6. Second difference: 2, so the coefficient of n^2 is 1.
2. Subtract n^2: 5-1=4, 7-4=3, 11-9=2, 17-16=1.
3. The remainders 4,3,2,1 go down by 1 each time, so their nth term is -n+5.
4. The nth term is n^2-n+5.

The second difference is twice the coefficient of the squared term.

1. For an^2+bn+c, the constant second difference is 2a.
2. So 2a=6 and a=3.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 7,13,19. Second difference: 6, so the coefficient of n^2 is 6\div2=3.
2. The aligned table shows the terms and their constant second difference.

3. Subtract 3n^2 (that is 3,12,27,48): the remainders are 2,0,-2,-4.
4. The remainders go down by 2, starting at 2, so their nth term is -2n+4.
5. The nth term is 3n^2-2n+4.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 7,5,3. Second difference: -2, so the coefficient of n^2 is -2\div2=-1.
2. The aligned table shows the terms and their constant second difference.

3. Subtract -n^2, which means adding n^2: 9+1=10, 16+4=20, 21+9=30, 24+16=40.
4. The remainders 10,20,30,40 have nth term 10n.
5. The nth term is -n^2+10n.

Find the first differences, then halve the constant second difference to find the coefficient of the squared term.

1. First differences: 2,3,4,5. Second difference: 1, so the coefficient of n^2 is \frac12.
2. The aligned table shows the terms and their constant second difference.

3. Subtract \frac12n^2 (that is 0.5,2,4.5,8,12.5): the remainders are 2.5,3,3.5,4,4.5.
4. The remainders go up by 0.5, starting at 2.5, so their nth term is \frac12n+2.
5. The nth term is \frac12n^2+\frac12n+2, which can also be written \frac{n^2+n}{2}+2.

The first differences must increase by the same amount.

1. The first two differences are 4 and 6, so the next differences must be 8 and 10.
2. The aligned table shows the terms and their constant second difference.

3. The missing term is 12+8=20. This checks because 30-20=10.

Find the nth term, or keep the pattern of the differences going.

1. First differences: 6,8,10,12. Second difference: 2, so the coefficient of n^2 is 1.
2. Subtract n^2: 4-1=3, 10-4=6, 18-9=9, 28-16=12, 40-25=15. These remainders are 3n.
3. So the nth term is n^2+3n.
4. The 10th term is 10^2+3\times10=130.

Set the term formula equal to the given value, then use the positive whole-number solution.

1. Solve n^2+n=182, or n^2+n-182=0.
2. Factorise: (n-13)(n+14)=0.
3. A term number is positive, so reject -14. The value occurs in term 13.

Think about whether every term must be odd or even.

1. n^2 is an integer, so 2n^2 is even.
2. Adding 1 makes every term odd. Since 150 is even, it cannot be a term.

Use the actual position numbers when subtracting the squared part.

1. The two first differences shown are 15 and 21, so the second difference is 6. The squared part is 3n^2.
2. The aligned table shows the terms and their constant second difference.

3. At positions 2,3,4, this part gives 12,27,48. Each is 1 below the required term.
4. The rule is 3n^2+1.

Substitute each position to form two linear equations.

Part (a)

1. First term (n=1): a+b+4=8, so a+b=4.
2. Second term (n=2): 4a+2b+4=18, so 4a+2b=14.
3. Double the first equation: 2a+2b=8. Subtract it from the second: 2a=6, so a=3.

Part (b)

1. Use a+b=4 with a=3: b=1.
2. Check: the nth term 3n^2+n+4 gives 8 and 18 for n=1 and n=2.

Find the term formula, then check the two positions on either side of the target.

1. First differences: 9,13,17. Second difference: 4, so the coefficient of n^2 is 2.
2. Subtract 2n^2 (that is 2,8,18,32): the remainders 4,7,10,13 have nth term 3n+1. So the nth term is 2n^2+3n+1.
3. Try terms near 100: the 6th term is 2\times36+18+1=91 and the 7th term is 2\times49+21+1=120.
4. The terms keep increasing, so the first term greater than 100 is term 7.

Replace every occurrence of the position by the next position before subtracting.

1. Replace every n with n+1: T_{n+1}=3(n+1)^2-5(n+1)+2.
2. Expand: 3(n^2+2n+1)-5n-5+2=3n^2+6n+3-5n-3, which simplifies to 3n^2+n.
3. Subtract, using a bracket for T_n: T_{n+1}-T_n=3n^2+n-(3n^2-5n+2).
4. The 3n^2 terms cancel: n+5n-2=6n-2.

Equate the two expressions, then discard any solution that is not a positive whole-number position.

1. Set 3n^2+2=n^2+5n, giving 2n^2-5n+2=0.
2. Factorise: (2n-1)(n-2)=0, so n=\frac12 or n=2.
3. Only 2 is a positive integer. Both sequences have value 14 at that position.

Use the constant second difference to build the term formula.

Part (a)

1. First differences: 7,11,15. Second difference: 4, so the coefficient of n^2 is 2.
2. Subtract 2n^2 (that is 2,8,18,32): the remainders are 0,1,2,3, with nth term n-1.
3. The nth term is 2n^2+n-1.

Part (b)

1. Solve 2n^2+n-1=405, which rearranges to 2n^2+n-406=0.
2. Factorise: (n-14)(2n+29)=0, so n=14 or n=-14.5.
3. A term number is a positive whole number, so it is term 14.

Find the centre of the completed square, then check the neighbouring whole-number positions.

1. Complete the square: n^2-7n+5=\left(n-\frac72\right)^2-\frac{29}{4}.
2. This is smallest when n=\frac72, but n must be a whole number.
3. The whole numbers nearest to \frac72 are 3 and 4, both \frac12 away, so they give the same value.
4. Check: the 3rd term is 9-21+5=-7 and the 4th term is 16-28+5=-7, while the 2nd and 5th terms are both -5.
5. The smallest value, -7, is at term numbers 3 and 4.

Use the second difference to fix the squared coefficient before using the two given terms.

1. The coefficient of n^2 is half the second difference, 4\div2=2, so the nth term is 2n^2+bn+c.
2. First term (n=1): 2+b+c=5, so b+c=3.
3. Fifth term (n=5): 50+5b+c=57, so 5b+c=7.
4. Subtract the equations: 4b=4, so b=1 and c=2.
5. The nth term is 2n^2+n+2.