CHAPTER 22 THE BINOMIAL SERIES
EXERCISE 93 Page 195
1. Use Pascal's triangle to expand (x – y)7
From Pascal’s triangle on page 195 of the textbook,
(a + x)6 =
a 6 + 6a 5 x + 15a 4 x 2 + 20a 3 x3 + 15a 2 x 4 + 6ax5 + x 6
Thus, (a + x)7 =
a 7 + 7 a 6 x + 21a 5 x 2 + 35a 4 x3 + 35a 3 x 4 + 21a 2 x5 + 7 ax 6 + x 7
and
( x − y )7 = x 7 + 7 x 6 (− y ) + 21x5 (− y ) 2 + 35 x 4 (− y )3 + 35 x 3 (− y ) 4 + 21x 2 (− y )5 + 7 x(− y )6 + (− y )7
i.e.
( x − y )7 =x 7 − 7 x 6 y + 21x 5 y 2 − 35 x 4 y 3 + 35 x 3 y 4 − 21x 2 y 5 + 7 xy 6 − y 7
2. Expand (2a + 3b) 5 using Pascal’s triangle.
From page 195 of the textbook,
(a + x)
5
=
a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x3 + 5a x 4 + x5
Thus, replacing a with 2a and x with 3b gives:
( 2a + 3b )
5
=
( 2a ) + 5 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 5 ( 2a )( 3b ) + ( 3b )
5
4
3
2
2
3
4
5
= 32a 5 + 240a 4b + 720a 3b 2 + 1080a 2b3 + 810ab 4 + 243b5
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© 2014, John Bird
EXERCISE 94 Page 197
1. Use the binomial theorem to expand (a + 2x)4
( a + 2x)
= a 4 + 4a 3 ( 2 x ) +
4
( 4 )( 3) a 2
2!
( 2x)
2
+
( 4 )( 3)( 2 ) a
3!
(2x) + (2x)
3
4
= a 4 + 8a 3 x + 24a 2 x 2 + 32 a x 3 + 16 x 4
2. Use the binomial theorem to expand (2 – x)6
(2 − x)
6
= 26 + 6(2)5 ( − x ) +
( 6 )( 5) (2)4
2!
(−x)
2
+
( 6 )( 5)( 4 ) (2)3
3!
(−x)
+
3
+
(6)(5)(4)(3) 2
(2) (− x) 4
4!
(6)(5)(4)(3)(2)
(2)(− x)5 + (− x)6
5!
= 64 − 192 x + 240 x 2 − 160 x3 + 60 x 4 − 12 x5 + x 6
3. Expand (2x – 3y)4
( 2 x − 3 y=
) ( 2x)
4
4
+ 4 ( 2 x ) ( −3 y ) +
3
( 4 )( 3)
2!
( 2 x ) ( −3 y )
2
2
+
( 4 )( 3)( 2 )
3!
( 2 x )( −3 y ) + ( −3 y )
3
4
= 16 x 4 − 96 x3 y + 216 x 2 y 2 − 216 xy 3 + 81 y 4
2

4. Determine the expansion of  2x + 
x

5
2

 2x +  =
x

5
( 5)( 4 ) 2 x 3  2  + ( 5)( 4 )( 3) 2 x 2  2 
2
( 2 x ) + 5 ( 2 x )   +
( ) 
( ) 
2!
3!
x
x
x
2
5
3
4
( 5)( 4 )( 3)( 2 )
+
4!
= 32 x5 + 160 x 3 + 320 x +
4
2
2
( 2 x )   +  
x x
5
320 160 32
+
+
x
x3 x5
5. Expand (p + 2q)11 as far as the fifth term.
346
© 2014, John Bird
(p + 2q)11 = p11 + 11 p10 ( 2q ) +
(11)(10 ) p9
2!
( 2q )
2
+
(11)(10 )( 9 ) p8
3!
( 2q )
3
+
(11)(10)(9)(8) 7
p (2q ) 4
4!
= p11 + 22 p10 q + 210 p 9 q 2 + 1320 p8 q 3 + 5280 p 7 q 4
13
q

6. Determine the sixth term of  3 p + 
3

The 6th term of
(a + x)
n
n ( n − 1)( n − 2 ) to (r − 1) terms
is
( r − 1)!
13
q

Hence, the 6th term of  3 p +  is:
3

a n −( r −1) x r −1
(13)(12 )(11)(10 )( 9 )
5!
(3 p )
13−5
5
8q
q
  = 1287 ( 3 p )  
3
3
5
= 34749 p8 q 5
7. Determine the middle term of (2a – 5b)8
With power 8 there will be 9 terms in the series. The middle term will be the fifth term
(2a – 5b)8
= (2a )8 + 8(2a)7 ( −5b ) +
(8)( 7 ) (2a)6
2!
( −5b )
2
+
(8)( 7 )( 6 ) (2a)5
3!
( −5b )
3
+
(8)(7)(6)(5)
(2a ) 4 (−5b) 4
4!
(8)(7)(6)(5)
(2a ) 4 (−5b) 4 = 700000 a 4b 4
4!
Hence, the fifth term is:
8. Use the binomial theorem to determine, correct to 4 decimal places:
(a) (1.003)8
(a) (1.003) =
8
(b) (0.98)7
(1 + 0.003)
8

(8)( 7 ) (0.003)2 + (8)( 7 )( 6 ) (0.003)3 + ...
= 1 + 8(0.003) +

2!
3!


= (1 + 0.024 + 0.000252 + 0.000001512 + ...) =
1.0242535...
= 1.0243, correct to 4 decimal places.
347
© 2014, John Bird
(b)
( 0.98)
7
=
1 + 7 ( −0.02 ) +
(1 − 0.02 ) =
7
( 7 )( 6 )
2!
( −0.02 )
2
+
( 7 )( 6 )( 5)
3!
( −0.02 )
3
+
( 7 )( 6 )( 5)( 4 )
4!
( −0.02 )
4
+ ...
= 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 – …
= 0.8681, correct to 4 decimal places
9. Evaluate (4.044)6 correct to 2 decimal places.
( 4.044 )
6
=
( 4 + 0.044 )
(1 + 0.011)
6
6
6
  0.044  
6
=+
46 (1 0.011)
  =+
 4 1
4 
 
=
1 + 6 ( 0.011) +
( 6 )( 5)
2!
( 0.011)
2
+
( 6 )( 5)( 3)
3!
( 0.011)
3
+
( 7 )( 6 )( 5)( 4 )
( 0.011)
4
4!
(7)(6)(5)(4)(3)
+
(0.011)5 + ...
5!
= 1 + 0.066 + 0.001815 + 0.00002662 + 0.000000512435 + 0.000000003382071
= 1.067842136...
Hence, (4.044)6 = 46 (1 + 0.011) = 46 (1.067842136...) = 4373.88, correct to 2 decimal places
6
348
© 2014, John Bird
EXERCISE 95 Page 199
1. Expand
1
in ascending powers of x as far as the term in x3, using the binomial theorem.
(1 − x)
State in each case the limits of x for which the series is valid.
( −1)( −2 ) − x 2 + ( −1)( −2 )( −3) − x 3 + ...
1
−1
= (1 − x ) = 1 + (−1)(− x) +
( )
( )
2!
3!
(1 − x )
= 1 + x + x 2 + x 3 + ...
2. Expand
x <1
and
1
in ascending powers of x as far as the term in x3, using the binomial theorem.
(1 + x) 2
State in each case the limits of x for which the series is valid.
1
(1 + x )
1 2x +
=(1 + x ) =−
−2
2
( −2 )( −3)
2!
( x)
2
= 1 − 2 x + 3 x 2 − 4 x 3 + ...
3. Expand
+
( −2 )( −3)( −4 )
3!
and
( x)
3
+ ...
x <1
1
in ascending powers of x as far as the term in x3, using the binomial theorem.
(2 + x)3
State in each case the limits of x for which the series is valid.
−3
2
3


1
−3
 x
 x  ( −3)( −4 )  x  ( −3)( −4 )( −5 )  x 
−
3
−
3
= ( 2 + x ) = 2 1 +  = 2 1 − 3   +
+
+ ...




(2 + x)3
2!
3!
 2
2
2
2


The series is true provided
x
<1
2
3 2 5 3
 3

1 − 2 x + 2 x − 4 x + ...


=
1
23
=
1 3
3
5

1 − x + x 2 − x 3 + ...

8 2
2
4

i.e.
x <2
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© 2014, John Bird
2 + x in ascending powers of x as far as the term in x3, using the binomial theorem.
4. Expand
State in each case the limits of x for which the series is valid.
  x 
 2 1 + 2   =

 
2+ x =
 x
2 1 +  =
 2
1/2
 x
2 1 + 
 2
Using the expansion of (1 + x)n,
1/2
 x
2 1 + 
 2
2[
=
=
2
3
]
 x x 2 x3

2 1 + − +
− ... 
 4 32 128

This is valid when
5. Expand
 1   x  (1/ 2)(−1/ 2)  x  (1/ 2)(−1/ 2)(−3 / 2)  x 
   +
  +
  + ...
2!
3!
 2  2 
2
2
x
< 1 i.e.
2
x
< 2 or –2 < x < 2
1
in ascending powers of x as far as the term in x3, using the binomial theorem.
1 + 3x
State in each case the limits of x for which the series is valid.
1
= (1 + 3 x )
1 + 3x
= 1−
1
−
2
 1  3 
 1  3  5 
−  − 

 −  −  − 
2
3
2  2 
 1

= 1 +  −  (3 x) +
( 3x ) +  2   2   2  ( 3x ) ...
2!
3!
 2
3
27 2 135 3
x+
x −
x as far as the term in x3
2
8
16
The series is true provided 3x < 1
i.e.
x <
1
3
6. Expand (2 + 3x)–6 to three terms. For what values of x is the expansion valid?
−6
2

 3x 
 3 x  ( −6 )( −7 )  3 x  
−
6
2 1 + (−6)   +
( 2 + 3x=
) 2 1 + =

   to three terms
2 
2!

 2 
 2  

−6
−6
=
1
26
189 2
1 
189 2



1 − 9 x + 4 x + ... = 64 1 − 9 x + 4 x + ...


350
© 2014, John Bird
The series is true provided
3x
<1
2
x <
i.e.
2
3
7. When x is very small show that:
(a)
(a)
1
(1 − x)
(1 − x)
2
1
(1 − x ) (1 − x )
2
≈1+
= (1 − x )
5
x
2
−2
(b)
(1 − x )
−
1
2
(1 − 2 x)
≈ 1 + 10x
(1 − 3 x) 4
(c)
3
1 + 5x
19
≈1+
x
6
1− 2x
 x
≈ (1 + 2 x ) 1 + 
 2
≈ 1+
x
+ 2 x ignoring the x 2 term and above
2
≈ 1+
5
x
2
(1 − 2 x ) =
−4
(1 − 2 x )(1 − 3x ) ≈ (1 − 2 x )(1 + 12 x )
4
(1 − 3x )
(b)
≈ 1 + 12x – 2x ignoring the x 2 term and above
≈ 1 + 10x
1 + 5x
(c)
3
(1 − 2 x )
1
= (1 + 5 x ) 2 (1 − 2 x )
−
1
3
 5  2 
≈ 1 + x 1 + x 
 2  3 
2
5
x + x ignoring the x 2 term and above
3
2
19
 5 2 15 + 4 19 
≈ 1+ x
=
 +=

6
6
6
2 3
≈ 1+
8. If x is very small such that x2 and higher powers may be neglected, determine the power series for
x+ 4 3 8− x
5 (1 + x )3
x+4
5
3
8− x
(1 + x )
3
1
1
2
1
3
=
( 4 + x ) (8 − x ) (1 + x )
≈
−
3
5
1
3
−
 x 2 1  x 3
=
4 1 +  8 3 1 −  (1 + x ) 5
 4
 8
1
2
( 4 )( 8 ) 1 + 12  4x  − 13  8x  − 53 ( x ) 
3
ignoring the x 2 term and above
 x x 3x 
 62 x 
 15 x − 5 x − 72 x 
≈ 4 1 + − −  ≈ 4 1 +
 ≈ 4 1 −

120


 8 24 5 
 120 
351
© 2014, John Bird
31
 31x 
≈ 4 1 −
x
 =4–
60 
15

9. Express the following as power series in ascending powers of x as far as the term in x2. State in
each case the range of x for which the series is valid.
(a)
(a)
 1− x 


 1+ x 
(1 + x) 3 (1 − 3 x) 2
(b)
 1− x 

 = (1 − x ) (1 + x )
 1+ x 
1
2
1
−
2
(1 + x 2 )


 1  1 
 1  3  
−
−
 x  2  − 2 


x  2   2  2 
2




≈ 1 − +
x 
( − x )  1 − +
2!
2!
 2
 2


 

as far as the term in x 2
x 3x 2 x x 2 x 2
 x x 2  x 3 x 2 
= 1 − − 1 − +
≈
−
+
− + −
1

2 8
2 4 8
 2 8  2 8 
≈ 1− x +
The series is valid if
(b)
(1 + x ) 3 (1 − 3x )
(1 + x 2 )
x2
as far as the term in x 2
2
x <1
2
2
=
(1 + x )(1 − 3x ) 3 (1 + x 2 )
−
1
2


 2  1 
  − 


x2
2
3
3
≈ (1 + x ) 1 − 2 x +   
( −3x ) + ... 1 − + ...  as far as the term in x 2
2!
2





 x2 
≈ (1 + x )(1 − 2 x − x 2 ) 1 − 
2 

 x2 
≈ (1 − 2 x − x 2 + x − 2 x 2 ) 1 −  as far as the term in x 2
2 

x2
 x2 
≈ (1 − x − 3 x 2 ) 1 −  ≈ 1 − − x − 3 x 2 neglecting x3 terms and above
2 
2

≈ 1− x −
The series is valid provided
7 2
x
2
3x < 1
i.e.
352
x <
1
3
© 2014, John Bird
EXERCISE 96 Page 201
1. Pressure p and volume v are related by pv3 = c, where c is a constant. Determine the approximate
percentage change in c when p is increased by 3% and v decreased by 1.2%.
New pressure = 1.03p = (1 + 0.03)p and new volume = 0.988v = (1 – 0.012)v
New value of c = (1 + 0.03)p(1 – 0.012) 3v3
= pv3(1 + 0.03)(1 – 0.012) 3
≈ pv3(1 + 0.03 – (3)0.012)
≈ pv3(1 + 0.03 – 0.036) ≈ pv3(1 – 0.006)
i.e.
99.4% of the original value of c
Thus, the approximate percentage change in c is a reduction of 0.6%
2. Kinetic energy is given by
1 2
mv . Determine the approximate change in the kinetic energy when
2
mass m is increased by 2.5% and the velocity v is reduced by 3%.
New mass = 1.025m = (1 + 0.025)m
New kinetic energy =
and
new velocity = 0.97v = (1 – 0.03)v
1
1
(1 + 0.025)m (1 − 0.03) 2 v 2 ≈ mv 2 (1 + 0.025)(1 − 0.06)
2
2
≈
i.e.
1 2
1 2
mv (1 + 0.025 − 0.06) =
mv (0.965)
2
2
96.5% of the original kinetic energy
Thus, the approximate change in kinetic energy is a reduction of 3.5%
3. An error of +1.5% was made when measuring the radius of a sphere. Ignoring the products of
small quantities, determine the approximate error in calculating (a) the volume and (b) the surface
area.
(a) Volume of sphere, V =
4 3
πr
3
353
© 2014, John Bird
New volume =
4
4
4
4
π (1.015r )3 = π r 3 (1 + 0.015)3 ≈ π r 3[1 + 3(0.015)] = π r 3 (1 + 0.045)
3
3
3
3
= 1.045V
i.e.
the volume has increased by 4.5%
(b) Surface area of sphere, A = 4π r 2
New surface area = 4π (1 + 0.015 ) r 2 ≈ 4π r 2 [1 + 2(0.015)] = 4π r 2 (1 + 0.03)
2
= 1.03A
i.e.
the surface area has increased by 3.0%
4. The power developed by an engine is given by I = kPLAN, where k is a constant. Determine the
approximate percentage change in the power when P and A are each increased by 2.5% and L and
N are each decreased by 1.4%.
I = kPLAN
New power = k (1 + 0.025) P (1 − 0.014) L(1 + 0.025) A(1 − 0.014) N
≈ kPLAN (1 + 0.025 − 0.014 + 0.025 − 0.014)
= kPLAN (1 + 0.022)
i.e.
102.2% of the original power
Thus, the approximate percentage change in power is an increase of 2.2%
5. The radius of a cone is increased by 2.7% and its height reduced by 0.9%. Determine the
approximate percentage change in its volume, neglecting the products of small terms.
Volume of cone =
New volume =
1 2
πr h
3
1
1
2
π (1 + 0.027 ) r 2 (1 − 0.009 ) h ≈ π r 2 h (1 + (2)0.027 − 0.009 )
3
3
≈
i.e.
1 2
π r h (1 + 0.045 )
3
104.5% of the original volume
Thus, the approximate percentage change in volume is an increase of 4.5%
354
© 2014, John Bird
6. The electric field strength H due to a magnet of length 2l and moment M at a point on its axis
distance x from the centre is given by:
H=
M 1
1 
−


2l  ( x − l ) 2 ( x + l ) 2 
Show that if l is very small compared with x, then H ≈
H
=
M
2l
 1

2
 ( x − l )


1  M 
1
−
=

2
( x + l )  2l  x 2 1 −
 
≈
2M
x3


1
 M 
−
=
 1−
2
2
2l 
2
x
l
l



x 2 1 +  

x
 x  
−2
l

 − 1 +
x

l

x
−2



M  2l   2l   M  4l  2M
1 +  − 1 −   ≈ 2   ≈ 3
x
2 x 2l 
x 
x  2 x l  x 
7. The shear stress t in a shaft of diameter D under a torque T is given by: τ =
kT
π D3
Determine the approximate percentage error in calculating τ if T is measured 3% too small and D
1.5% too large.
New value of T = (1 – 0.03)T
Hence, new shear stress =
≈
and
new value of D = (1 + 0.015)D
k (1 − 0.03)T
kT 
−3
=
1 − 0.03)(1 + 0.015 ) 
(
3
3
3

π (1 + 0.015) D π D 
kT
kT
(1 − 0.03)(1 − 0.045 )  ≈
[1 − 0.03 − 0.045]
3
πD
π D3
≈
i.e.
kT
τ (1 − 0.075)
(1 − 0.075) =
π D3
the new torque has decreased by 7.5%
355
© 2014, John Bird
8. The energy W stored in a flywheel is given by: W = kr5N2, where k is a constant, r is the radius
and N the number of revolutions. Determine the approximate percentage change in W when r is
increased by 1.3% and N is decreased by 2%.
W = kr5N2
New energy = k(1 + 0.013) 5r5 (1 – 0.02) 2N2
≈ kr5N2(1 + (5)0.013 – (2)0.02)
≈ kr5N2(1 + 0.065 – 0.04) ≈ kr5N2(1 + 0.025)
i.e.
102.5% of the original volume
Thus, the approximate percentage change in energy is an increase of 2.5%
9. In a series electrical circuit containing inductance L and capacitance C the resonant frequency is
given by: f r =
1
2π LC
. If the values of L and C used in the calculation are 2.6% too large and
0.8% too small, respectively, determine the approximate percentage error in the frequency.
New value of inductance = (1 + 0.026)L
Hence, new resonant frequency =
=
≈
and
new value of capacitance = (1 – 0.008)C
1
1
=
2π (1 + 0.026) L(1 − 0.008)C 2π (1 + 0.026) L (1 − 0.008)C
1
1
1
1
1
−
(1 + 0.026 ) 2 L− 2 (1 − 0.008)− 2 C − 2
2π
1
1
(1 − 0.013)(1 + 0.004 )  ≈
(1 − 0.013 + 0.004 )
π
LC
2
2π L C
1
2
1
2
≈ f r (1 − 0.009 )
i.e.
the new resonant frequency is 0.9% smaller
10. The viscosity η of a liquid is given by: η =
kr 4
, where k is a constant. If there is an error in r of
νl
+2%, in υ of +4% and I of –3%, what is the resultant error in η?
356
© 2014, John Bird
new value of ν = (1 + 0.04)ν and new value of l = (1 – 0.03) l
New value of r = (1 + 0.02)r,
Hence, new value of viscosity =
i.e.
k (1 + 0.02) 4 r 4
kr 4
4
−1
−1
= (1 + 0.02 ) (1 + 0.04 ) (1 − 0.03) 


(1 + 0.04)ν (1 − 0.03) l ν l
≈
kr 4
(1 + 0.08 )(1 − 0.04 )(1 + 0.03) 
νl 
≈
kr 4
(1 + 0.08 − 0.04 + 0.03) ≈ η (1 + 0.07 )
νl
the viscosity increases by 7%
11. A magnetic pole, distance x from the plane of a coil of radius r, and on the axis of the coil, is
subject to a force F when a current flows in the coil. The force is given by: F =
kx
( r 2 + x2 )
5
,
where k is a constant. Use the binomial theorem to show that when x is small compared to r,
kx 5 kx 3
−
r 5 2r 7
then F ≈
F=
kx
(r
2
+x
2
)
5
=
kx
( r 2 + x2 )
5
2
= kx ( r 2 + x 2 )
−
5
− 

5
2
2
x
− 

= kx ( r 2 ) 2 1 +  

 r2  


5
2
 x2 
= kxr − 5 1 + 
 r2 
−
5
2
≈
kx   5  x 2 
1 +  −   when x is small compared to r
r5   2  r2 
≈
kx  5 x 2  kx kx  5 x 2 
− 
1 −
 ≈

r 5  2r 2  r 5 r 5  2r 2 
≈
kx 5kx 3
−
r 5 2r 7
12. The flow of water through a pipe is given by: G =
( 3d )
5
H
L
. If d decreases by 2% and H by
1%, use the binomial theorem to estimate the decrease in G.
357
© 2014, John Bird
New value of d = (1 – 0.02)d
and
( 3d ) H
Hence, new value of G = =
5
L
5
new value of H = (1 – 0.01)H
5
1
35 d 2 H 2
=
1
L2
5
5
1
1
35 (1 − 0.02) 2 d 2 (1 − 0.01) 2 H 2
1
L2
1
35 d 2 H 2  5
 1

≈
1 − 2 (0.02)  1 − 2 (0.01)  
1



L2
≈
(3d )5 H
(3d )5 H
(1 − 0.05 )(1 − 0.005 )  ≈
(1 − 0.05 − 0.005)
L
L
≈ G (1 − 0.055)
i.e.
the flow G has decreased by 5.5%
358
© 2014, John Bird