Chapter 12 Sequences; Induction; the Binomial Theorem

Section 12.2: Arithmetic Sequences
Section 12.2
1. arithmetic
2. False; the sum of the first and last terms equals
twice the sum of all the terms divided by the
number of terms.
3. d = sn
−sn−1
= ( n+ 4) −( n− 1+ 4) = ( n+ 4) − ( n+
3)
= n+ 4−n− 3=
1
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
s1 = 1+ 4= 5, s2 = 2+ 4= 6, s3
= 3+ 4=
7,
s = 4+ 4=
8
4.
5.
6.
4
d = s −s
n
n−1
( ) ( )
= ( n−5) −( n−1− 5) = n−5 − n−6
= n−5− n+ 6=
1
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
s1 = 1− 5 =− 4, s2 = 2 − 5 =− 3, s3
= 3 − 5 =−2,
s = 4− 5= −1
4
d = an
−an−1
= ( 2n−5 ) −(2( n−1) −5)
= ( 2n−5) −( 2n−2−5)
= 2n−5− 2n+ 7 = 2
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
a1 = 21 ⋅ − 5= − 3, a2
= 22 ⋅ − 5= −1,
a = 23 ⋅ − 5= 1, a = 24 ⋅ − 5=
3
3 4
d = bn
−bn−1
= ( 3n+ 1 ) −(3( n− 1) + 1)
= ( 3n+ 1) −( 3n− 3+
1)
= 3n+ 1− 3n+ 2=
3
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
b1 = 31 ⋅ + 1= 4, b2
= 32 ⋅ + 1=
7,
b = 3⋅ 3+ 1= 10, b = 3⋅ 4+ 1=
13
3 4
7.
8.
d = c − c
n
n−1
( n)
n
( 6 2n) ( 6 2n
2)
= 6−2 −(6−2( −1))
= − − − +
= 6−2n− 6+ 2n− 2=−2
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
c = 6−2⋅ 1= 4, c = 6−2⋅ 2=
2,
c
1 2
= 6−2⋅ 3= 0, c = 6−2⋅ 4=−2
3 4
d = d − d
n
n−1
( n)
n
( 4 2n) ( 4 2n
2)
= 4−2 −(4−2( −1))
= − − − +
= 4−2n− 4+ 2n− 2=−2
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
d = 4−2⋅ 1= 2, d = 4−2⋅ 2=
0,
d
1 2
= 4−2⋅ 3= − 2, d = 4−2⋅ 4= −4
3 4
9. d = tn
− tn−1
⎛1 1 ⎞ ⎛1 1 ⎞
= ⎜ − n⎟−⎜ − ( n−
1) ⎟
⎝2 3 ⎠ ⎝2 3 ⎠
⎛1 1 ⎞ ⎛1 1 1⎞
= ⎜ − n⎟−⎜ − n+
⎟
⎝2 3 ⎠ ⎝2 3 3⎠
1 1 1 1 1 1
= − n− + n− = −
2 3 2 3 3 3
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
1 1 1 1 1 1
t1 = − ⋅ 1 = , t2
= − ⋅ 2 =− ,
2 3 6 2 3 6
1 1 1 1 1 5
t3 = − ⋅ 3 =− , t4
= − ⋅ 4=−
2 3 2 2 3 6
10. d = tn
− tn−1
⎛2 1 ⎞ ⎛2 1 ⎞
= ⎜ + n⎟− ⎜ + ( n−
1) ⎟
⎝3 4 ⎠ ⎝3 4 ⎠
⎛2 1 ⎞ ⎛2 1 1⎞
= ⎜ + n⎟− ⎜ + n−
⎟
⎝3 4 ⎠ ⎝3 4 4⎠
2 1 2 1 1 1
= + n− − n+ =
3 4 3 4 4 4
The difference between consecutive terms is
constant, therefore the sequence is arithmetic.
2 1 11 2 1 7
t1 = + ⋅ 1 = , t2
= + ⋅ 2 = ,
3 4 12 3 4 6
2 1 17 2 1 5
t3 = + ⋅ 3 = , t4
= + ⋅ 4=
3 4 12 3 4 3
1247
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