Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

62 Higher Engineering Mathematics



  1. Expand(p+ 2 q)^11 as far as the fifth term.



p^11 + 22 p^10 q+ 220 p^9 q^2

+ 1320 p^8 q^3 + 5280 p^7 q^4




  1. Determine the sixth term of


(
3 p+

q
3

) 13
.

[34749p^8 q^5 ]


  1. Determine the middle term of( 2 a− 5 b)^8.
    [700000a^4 b^4 ]

  2. Use the binomial theorem to determine, cor-
    rect to 4 decimal places:
    (a)( 1. 003 )^8 (b)( 1. 042 )^7
    [(a) 1.0243 (b) 1.3337]

  3. Use the binomial theorem to determine, cor-
    rect to 5 significant figures:
    (a)( 0. 98 )^7 (b)( 2. 01 )^9
    [(a) 0.86813 (b) 535.51]

  4. Evaluate( 4. 044 )^6 correct to 3 decimal places.
    [4373.880]


7.4 Further worked problems on the


binomial series


Problem 11.
(a) Expand
1
( 1 + 2 x)^3

in ascending powers ofxas

far as the term inx^3 , using the binomial series.
(b) State the limits ofxfor which the expansion
is valid.

(a) Using the binomial expansion of( 1 +x)n,where
n=−3andxis replaced by 2xgives:
1
( 1 + 2 x)^3

=( 1 + 2 x)−^3

= 1 +(− 3 )( 2 x)+

(− 3 )(− 4 )
2!

( 2 x)^2

+

(− 3 )(− 4 )(− 5 )
3!

( 2 x)^3 +···

= 1 − 6 x+ 24 x^2 − 80 x^3 +···

(b) The expansion is valid provided| 2 x|<1,

i.e. |x|<

1
2

or−

1
2

<x<

1
2

Problem 12.
(a) Expand

1
( 4 −x)^2

in ascending powers ofxas
far as the term inx^3 , using the binomial
theorem.
(b) What are the limits ofxfor which the expan-
sion in (a) is true?

(a)

1
( 4 −x)^2

=

1
[
4

(
1 −

x
4

)] 2 =

1

42

(
1 −

x
4

) 2

=

1
16

(
1 −

x
4

)− 2

Using the expansion of( 1 +x)n
1
( 4 −x)^2

=

1
16

(
1 −

x
4

)− 2

=

1
16

[
1 +(− 2 )

(

x
4

)

+

(− 2 )(− 3 )
2!

(

x
4

) 2

+

(− 2 )(− 3 )(− 4 )
3!

(

x
4

) 3
+···

]

=

1
16

(
1 +

x
2

+

3 x^2
16

+

x^3
16

+···

)

(b) The expansion in (a) is true provided

∣∣

x
4

∣∣
∣<1,

i.e.|x|< 4 or− 4 <x< 4

Problem 13.√ Use the binomial theorem to expand
4 +xin ascending powers ofxto four terms. Give
the limits ofxfor which the expansion is valid.


4 +x=

√[
4

(
1 +

x
4

)]

=


4

√(
1 +

x
4

)
= 2

(
1 +

x
4

)^12

Using the expansion of( 1 +x)n,

2

(
1 +

x
4

)^1
2

= 2

[
1 +

(
1
2

)(
x
4

)
+

( 1 / 2 )(− 1 / 2 )
2!

(x
4

) 2
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