Download Exercise 3: Quadratic sequences

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Exercise 3: Quadratic sequences
Problem 1:
Determine whether each of the following sequences is:
a linear sequence;
a quadratic sequence;
or neither.
1.
8; 17; 32; 53; 80; …
2.
3p 2 ; 6p 2 ; 9p 2 ; 12p 2 ; 15p 2 ; …
3.
1; 2,5; 5; 8,5; 13; …
4.
2; 6; 10; 14; 18; …
5.
5; 19; 41; 71; 109; …
6.
3; 9; 16; 21; 27; …
7.
2k; 8k; 18k; 32k; 50k; …
8.
2 12 ; 6; 10 12 ; 16; 22 12 ; …
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Answer 1:
First differences: = 9; 15; 21; 27
Second difference: = 6
1.
Quadratic sequence
First difference: = 3p2
2.
Linear sequence
3.
First differences: = 1,5; 2,5; 3,5; 4,5
Second difference: = 1
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Quadratic sequence
First difference: = 4
4.
Linear sequence
First differences: = 14; 22; 30; 38
Second difference: = 8
5.
Quadratic sequence
6. Neither
First differences: = 6k; 10k; 14k; 18k
Second difference: = 4k
7.
Quadratic sequence
First differences: = 3,5; 4,5; 5,5; 6,5
Second difference: = 1
8.
Quadratic sequence
Problem 2:
A quadratic pattern is given by
Tn = n 2 + bn + c
. Find the values of band c if the sequence starts
with the following terms:
−1 ; 2 ; 7 ; 14 ; …
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Answer 2:
Starting with the first term, we have n = 1 and T1
= −1 :
T1 = (1) 2 + b(1) + c
(−1) = 1 + b + c
−2 = b + c
For the second term, we use n = 2 and T2
= (2
2
=2 :
+ b(2) + c
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T2 = (2) 2 + b(2) + c
(2) = 4 + 2b + c
−2 = 2b + c
Now we must solve these equations simultaneously. We can do this by substitution, but here we will
show the solution using the 'elmination' method (which means subtracting one equation from the other
to cancel the c's).
−2 = 2b + c
−(−2 = b + c)
0= b
Finally, calculate the value of c. As usual for simultaneous equations, this means that we must
substitute the b = 0 into either of the equations we used above. Let's use the equation −2 = b + c .
b=0
−2 = b + c
−2 = (0) + c
−2 = c
The final answers are b = 0 and c = −2 .
NOTE: Now we know that the general term of the sequence is
check our answers. We know that T3
Tn =
T3 =
=
=
Tn = n 2 − 2 . We can use this to
= 7 . Substitute n = 3 into the general formula to check:
n2 − 2
2
(3) − 2
(9) + 0 − 2
7
Problem 3:
a 2 ; −a 2 ; −3a 2 ; −5a 2 ; …
are the first 4 terms of a sequence.
1. Is the sequence linear or quadratic? Motivate your answer.
2. What is the next term in the sequence?
3. Calculate T100 .
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Answer 3:
=−
2
−
2
= −2
2
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T2 − T1 = −a 2 − a 2 = −2a 2
T3 − T2 = −3a 2 − (−a 2 ) = −2a 2
T4 − T3 = −5a 2 − (−3a 2 ) = −2a 2
1.
This is an arithmetic sequence since there is a common difference of −2a 2 between consecutive
terms.
T5 = −5a 2 + (−2a 2 )
2.
= −7a 2
Tn =
∴ T100 =
=
∴ T100 =
3.
a + (n − 1)d
a 2 + (99)(−2a 2 )
a 2 − 198a 2
−197a 2
Problem 4:
Given
Tn = n 2 + bn + c
, determine the values of b and c if the sequence starts with the terms:
2 ; 7 ; 14 ; 23 ; …
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Answer 4:
Starting with the first term, we have n = 1 and T1
=2 :
T1 = (1) 2 + b(1) + c
(2) = 1 + b + c
1= b+c
For the second term, we use n = 2 and T2
=7 :
T2 = (2) 2 + b(2) + c
(7) = 4 + 2b + c
3 = 2b + c
Now we must solve these equations simultaneously. We can do this by substitution, but here we will
show the solution using the 'elimination' method (which means subtracting one equation from the
other to cancel the c's).
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3 = 2b + c
−(1 = b + c)
2= b
Finally, calculate the value of c. As usual for simultaneous equations, this means that we must
substitute the b = 2 into either of the equations we used above. Let's use the equation 1 = b + c .
b=2
1 = b+c
1 = (2) + c
−1 = c
The final answers are
b = 2 and c = −1 .
Tn = n 2 + 2n − 1 . We can use this
to check our answers. We know that T3 = 14 . Substitute n = 3 into the general formula to check:
NOTE: Now we know that the general term of the sequence is
Tn =
T3 =
=
=
n 2 + 2n − 1
(3) 2 + 2(3) − 1
(9) + 6 − 1
14
Problem 5:
The first term of a quadratic sequence is 4, the third term is 34 and the common second difference is
10. Determine the first six terms in the sequence.
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Answer 5:
Let T2 =
∴ T2 − T1 =
And T3 − T2 =
Second difference =
=
∴ 10 =
2x =
∴x=
x
x−4
34 − x
(T3 − T2 ) − (T2 − T1 )
(34 − x) − (x − 4)
38 − 2x
28
14
4; 14; 34; 64; 104; 154
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Problem 6:
A quadratic sequence has a second term equal to 1, a third term equal to −6 and a fourth term equal
to −14.
1. Determine the second difference for this sequence.
2. Hence, or otherwise, calculate the first term of the pattern.
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Answer 6:
1.
T3 − T2 =
=
T4 − T3 =
=
∴ Second difference =
2.
T1 = 1 + 6
=7
−6 − (1)
−7
−14 − (−6)
−8
−1
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