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Exercise 8.3 (Solutions)
Textbook of Algebra and Trigonometry for Class XI
Available online @ http://www.mathcity.org, Version: 3.6
Binomial Theorem when n is negative or fraction:
When n is negative or fraction and x < 1 then
n (n − 1) 2 n (n − 1)(n − 2) 3
(1 + x)n = 1 + n x +
x +
x + ...
2!
3!
Where the general term of binomial expansion is
n (n − 1)(n − 2)...( n − (r − 1) ) r
Tr +1 =
x
r!
Question # 1
Expand the following upto 4 times, taking the values of x such that the expansion in
each case is valid.
(i) (1 − x )
1
2
(ii) (1 + 2 x)
−1
(iii) (1 + x)
−
1
3
(iv) (4 − 3 x)
−1
(v) (8 − 2 x)
−1
(vi) (2 − 3x)
−2
(vii)
(1 − x )
2
(1 + x )
(viii)
(1 + 2 x )
(1 − x )
1
( 4 + 2x )2
(ix)
(2 − x)
Solution
(i)
1
2 2
(x) (1 + x − 2 x )
1
2 2
(xi) (1 − 2 x + 3 x )
11 
1  1  1

− 1
− 1 − 2 


1
2 2 
2 2  2
 (− x)3 + ...
(1 − x) = 1 + (− x) + 
(− x)2 + 
2
2!
3!
1 1
1  1  3 
− 

 −  − 
1
2  2  2 2  2  2 
=1− x +
x +
(− x3 ) + ...
2
2
3⋅ 2
1
1
1
= 1 − x − x 2 − x3 + ...
2
8
16
1
2
(ii)
Do yourself as above
(iii)
Do yourself as above
1
(iv)
1
1
1
  3x   2
 3x  2
 3x  2
(4 − 3 x ) =  4 1 −   = (4) 2  1 −  = 2 1 − 
4 
4 
4 


 
1
2
1
2

11 
1  1  1


 1  3 x  2  2 − 1  3 x  2 2  2 − 1 2 − 2   3 x 3

 −


 −
= 2 1 +  −  + 
+
+
...





2
4
2!
4
3!
4










FSc-I / Ex 8.3 - 2

1 1
1  1  3 

 3 x 2  − 2   9 x 2  2  − 2  − 2   27 x 3 






= 2 1 −
+

+
−
 + ...
8
2  16 
3⋅ 2


 64 


 3 x 1  9 x 2  1  27 x3 

= 2 1 − − 
−
+
...




8 8  16  16  64 


 3 x 9 x 2 27 x 3

= 2 1 −
−
−
+ ...
8 128 1024


3x 9 x 2 27 x3
=2− −
−
+ ...
4 64 512
−1
1
2
(v)
1 x 
 2x 
(8 − 2 x) = (8) 1 −  = 1 − 
8 
8 4 

(vi)
Do yourself
−1
−1
Now do yourself
(1 − x) −1
= (1 − x) −1 (1 + x) −2
(vii)
2
(1 + x)
(−1)(−1 − 1)
(−1)(−1 − 1)(−1 − 2)


= 1 + (−1)(− x) +
( − x) 2 +
(− x )3 + … 
2!
3!


(−2)(−2 − 1) 2 (−2)(−2 − 1)(−2 − 2) 3


× 1 + (−2)( x) +
( x) +
( x) + … 
2!
3!


(−1)(−2) 2 (−1)(−2)(−3)


= 1 + x +
(x ) +
(− x3 ) + … 
2
3⋅ 2


(−2)(−3) 2 (−2)(−3)(−4) 3


× 1 − 2 x +
( x) +
( x) + … 
2
3⋅ 2


2
3
2
= 1 + x + x + x + ................... × 1 − 2 x + 3x − 4 x3 +…
(
) (
)
= 1 + ( x − 2 x) + ( x 2 − 2 x 2 + 3 x 2 ) + ( x 3 − 2 x 3 + 3 x 3 − 4 x 3 ) + …
= 1 − x + 2 x 2 − 2 x3 +…
Do yourself as above
(viii)
1
(ix)
1
1
1
(4 + 2 x) 2
2x  2
 x
= (4 + 2 x) 2 (2 − x) −1 = (4) 2 1 +  (2)−1 1 − 
2− x
4 

 2
1
−1
1
−1
−1
1
x 2  x 
 x 2
 x
 x 2 1  x 

= (4) 1 +  (2)−1 1 −  = 2 1 +  1 −  = 1 +  1 − 
 2
 2
 2  2 2 
 2  2
1
2
−1
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FSc-I / Ex 8.3 - 3
1
−1
x 2  x 

= 1 +  1 − 
 2  2
11
11



 1 − 2 

 x 3
 1  x  2  2 − 1   x 2 2  2 − 1 
2
   + …
= 1 +   +
+

 
 
2!  2 
3!
2
 2  2 



2
3


 x  (−1)(−1 − 1)  x  (−1)(−1 − 1)(−1 − 2)  x 
×  1 + (−1)  −  +
−
+
−
+
…






2!
3!
 2
 2
 2


1 1
1  1  3 


 −   x3 
 x 2  − 2   x 2  2  − 2 

2
= 1 + +
+
+
…

 
 
2  4
3⋅ 2
8
 4



2
3


x (−1)(−2)  x  (−1)(−2)(−3)  x 
…
× 1 + +
+
−
+

 


2
3⋅ 2
 4
 8
 2


 

x x2
x3
x x 2 x3
= 1 + −
+
+ …  × 1 + + + + … 
8
 4 32 128
  2 4

2
x 2 x 2   x3 x3 x3 x3 
 x x  x
=1+  +  +  − + +  + 
−
+ +  +…
4   128 64 16 8 
 4 2   32 8
3x 11x 2 23 x3
=1+
+
+
+…
4
32
128
1
2 2
(1 + x − 2 x ) = (1 + ( x − 2 x ) )
(x)
2
1
2
11
11

 1 
− 1
− 1  − 2 


1
= 1 + ( x − 2 x 2 ) + 2  2  ( x − 2 x 2 ) 2 + 2  2  2  ( x − 2 x 2 )3 +…
2
2!
3!
1 1
1  1  3 
− 

 −  − 
1
2 2 2
2  2  2 
2
3
4
= 1 + ( x − 2x ) +
(x − 4x + 4x ) +
2
2
3⋅ 2
3
2
2
2 2
2 3
x + 3( x) (−2 x ) + 3( x)(−2 x ) − (2 x ) +…
(
(xi)
)
1
+ 4x ) + ( x
16
1
1
4
3
= 1 + ( x − 2 x 2 ) − ( x 2 − 4 x3
− 6 x 4 + 12 x5 − 8 x 6 ) +…
2
8
1
2
1
4
4
1
6
12
8
= 1 + x − x 2 − x 2 − x3 + x 4 + x3 − x 4 + x5 − x 6 +…
2
2
8
8
8
16
16
16
16
1
1
1
1
1
3
3
1
= 1 + x − x 2 − x 2 − x 3 + x 4 + x 3 − x 4 + x 5 − x 6 +…
2
8
2
2
16
8
4
8
1
9
9
= 1 + x − x 2 − x3 +…
2
8
16
Do yourself as above
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FSc-I / Ex 8.3 - 4
Question # 2
Use the Binomial theorem find the value of the following to three places of decimials.
1
1
(i) 99
(ii) ( 0.98 ) 2
(v) 4 17
(vi)
(iii) (1.03) 3
1
(vii) 3
998
1
(xi) 6
486
(ix)
7
8
5
31
(x) (0.998)
−
1
3
(iv)
3
(viii)
65
1
5
252
(xii) (1280)
1
4
Solution
1
1
2
1 2

99 = ( 99 ) = (100 − 1) = (100)  1 −

 100 
11



 1  1  2  2 − 1   1 2

= 10 1 +  −
+
−
+
…




2!  100 
 2  100 



1 1


− 



1
 1 
…
= 10 1 −
+ 2 2
+


2  10000 
 200



1


= 10 1 − 0.005 − ( 0.0001) + … 
8


= 10 (1 − 0.005 − 0.0000125 + …)
1
2
(i)
1
2
≈ 10 ( 0.9949875 ) = 9.949875
≈ 9.950
1
(ii)
(iii)
1
( 0.98) 2 = (1 − 0.02 ) 2
(1.03)
1
3
= (1 + 0.03)
1
3
Now do yourself
1
3
Now do yourself
1
3
1
1 3

= ( 64 − 1) = (64)  1 − 
 64 
1
3
3
65 = ( 65 )
(v)
4
1

17 = (17 ) = (16 − 1) = (16)  1 − 
 16 
(vi)
1 5

5
31 = ( 31) = ( 32 − 1) = (32)  1 − 
 32 
(iv)
1
4
1
5
1
4
1
5
1
4
1
5
1
4
Now do yourself
Now do yourself
1
Now do yourself
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FSc-I / Ex 8.3 - 5
(vii)
3
1
1
=
1 = ( 998 )
998 ( 998) 3
= (10
1
3 −3
)
1
−
3
1 

1 −

 500 
= (1000 − 2 )
−
1
−
3
2 

= (1000 )  1 −

 1000 
1
−
3
−
1
3
1
3
1
1


−  − − 1 
2


1 
1
 1  1 
3 3 
=   1 +  −  −
+
−
+
…




2!
 10    3  500 
 500 




1
4


−  − 


1 
1
 1 
3 3  
=   1 + 
+
+
…




2
 10    1500 
 250000 




2
 1 

=  1 + ( 0.0006667 ) + ( 0.000004 ) + … 
9
 10 

1
=   (1 + 0.0006667 + 0.00000089 + …)
 10 
1
≈   (1.00066759 ) = 0.100066759 ≈ 0.100 Answer
 10 
( viii)
1
1
1
1
1
−
−
9 
− 
5
5
5 1+
=
=
252
=
243
+
9
=
243
(
)
(
)
(
)
1


5
252 ( 252 ) 5
 243 
1
5 −5
= (3
)
1 

1 + 
 27 
1
(ix)
−
1
5
−
1
5
Now do yourself as above
1
7
7  7 2  1 2
=
=   = 1 − 
8 8  8
8
11

− 1
2

1 1 2 2  1
=1+  −  +
 −  +…
2 8
2!  8 
1 1
−
1 2  2   1 
=1− +
  +…
16
2  64 
1 1 1 
=1− −   +…
16 8  64 
1
1
=1− −
+…
16 512
= 1 − 0.0625 − 0.00195 + …
≈ 0.93555 ≈ 0.936
Answer
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FSc-I / Ex 8.3 - 6
(x)
(0.998)
−
1
3
= (1 − 0.002)
−
1
3
Now do yourself as above
1
1
1
1
1
−
−
243 
− 
6
6
6 1−
=
=
486
=
729
−
243
=
729
(
)
(
)
(
)
1


6
486 ( 486 ) 6
 729 
(xi)
1
6 −6
= (3
)
 1
1 − 
 3
1
4
−
1
4
1
6
−
1
6
Now do yourself
1
1
1
16  4
1 4

4 4
(1280) = (1296 − 16) = (1296 )  1 −
=
6
 ( ) 1 − 
 1296 
 81 
Now do yourself
(xii)
1
4
Question # 3
Find the coefficient of x n in the expansion of
2
1+ x)
1 + x2
(
(i)
(ii)
2
2
(1 + x )
(1 − x )
(1 + x)3
(iii)
(1 − x)2
2
(iv)
(1 + x )
3
(1 − x )
(v) (1 − x + x 2 − x 3 + …)
2
Solution
1 + x2
−2
= (1 + x 2 ) (1 + x )
(i)
2
(1 + x )
(−2)(−2 − 1) 2 (−2)(−2 − 1)(−2 − 2) 3


= (1 + x 2 ) 1 + (−2)( x) +
( x) +
( x) + … 
2!
3!


(−2)(−3) 2 (−2)(−3)(−4) 3


= (1 + x 2 ) 1 − 2 x +
x +
x + …
2
3⋅ 2


= (1 + x 2 )(1 − 2 x + 3x 2 − 4 x3 +…)
= (1 + x 2 )(1 + (−1)2 x + (−1)2 3 x 2 + (−1)3 4 x3 +…)
Following in this way we can write
1 + x2
= (1 + x 2 ) (1 + (−1)2 x + (−1) 2 3x 2 + (−1)3 4 x3 + … + (−1)n−2 (n − 1) x n−2 +
2
(1 + x )
(−1) n−1 ( n) x n−1 + (−1) n (n + 1) x n + …)
So taking only terms involving x n we get
(−1) n (n + 1) x n + (−1) n−2 (n − 1) x n
= (−1)n (n + 1) x n + (−1)n (−1)−2 (n − 1) x n
= (−1)n (n + 1) x n + (−1)n (n − 1) x n
∵ (−1) −2 = 1
= (n + 1 + n − 1)(−1) n x n = (2n)(−1) n x n
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FSc-I / Ex 8.3 - 7
Thus the coefficient of term involving x n is (2n)(−1)n
(ii)
Hint:
After solving you will get
2
(1 + x ) = 1 + 2 x + x 2 (1 + 2 x + 3x 2 + 4 x3 + … + (n − 1) x n−2 + (n) x n−1 + (n + 1) x n + …)
(
)
2
(1 − x )
Do yourself as above
(iii)
(1 + x)3
= (1 + x)3 (1 − x)−2
2
(1 − x)
(−2)(−2 − 1)
(−2)(−2 − 1)(−2 − 2)
3

= (1 + x ) 1 + (−2)(− x) +
(− x) 2 +
( − x )3 + … 
2!
3!


(−2)(−3) 2 (−2)(−3)(−4)
3

= (1 + x ) 1 + 2 x +
( x) +
(− x 3 ) + … 
2
3⋅ 2


= (1 + 3x + 3x 2 + x3 )(1 + 2 x + 3x 2 + 4 x3 +…)
Following in this way we can write
(1 + x)3
= 1 + 3x + 3x 2 + x3 (1 + 2 x + 3x 2 + 4 x3 + … + (n − 2) x n−3 + (n − 1) x n−2
2
(1 − x)
+(n) x n−1 + (n + 1) x n + …)
(
)
So taking only terms involving x n we have term
(n + 1) x n + 3(n) x n + 3(n − 1) x n + (n − 2) x n
= ( ( n + 1) + 3(n) + 3( n − 1) + ( n − 2) ) x n
= ( n + 1 + 3n + 3n − 3 + n − 2 ) x n
= ( 8n − 4 ) x n
Thus the coefficient of term involving x n is ( 8n − 4 ) .
2
(iv)
(1 + x ) = 1 + x 2 1 − x −3
( )( )
3
(1 − x )
(−3)(−3 − 1)
(−3)(−3 − 1)(−3 − 2)
2

= (1 + x ) 1 + (−3)(− x) +
( − x) 2 +
(− x)3 + ... 
2!
3!


(−3)(−4)
(−3)(−4)(−5)
2

= (1 + x ) 1 + (−3)(− x) +
(− x)2 +
(− x)3 + ... 
2
3⋅ 2


(3)(4) 2
(4)(5) 3


= (1 + 2 x + x 2 ) 1 + 3 x +
(x ) +
( x ) + ... 
2
2


(3)(4) 2 (4)(5) 3
 (2)(3)

= (1 + 2 x + x 2 ) 1 +
x+
x +
x + ... 
2
2
2


www.mathcity.org
FSc-I / Ex 8.3 - 8
Following in this way we can write
2
(1 + x ) = 1 + 2 x + x 2 1 + (2) (3) x + (3)(4) x 2 + (4) (5) x 3 + ...
(
) 
3
2
2
2
(1 − x )
(n − 1)(n) n−2 (n)(n + 1) n−1 (n + 1)(n + 2) n

x +
x +
x + ... 
2
2
2

n
So taking only terms involving x we have term
(n + 1)(n + 2) n
(n)(n + 1) n (n − 1)(n) n
x +2
x +
x
2
2
2
xn
= ( (n + 1)(n + 2) + 2(n)(n + 1) + (n − 1)(n) )
2
n
x
= ( n 2 + n + 2n + 2 + 2n 2 + 2n + n 2 − n )
2
n
x
xn
= ( 4n 2 + 4 n + 2 )
= 2 ( 2n 2 + 2n + 1)
2
2
2
n
= 2n + 2n + 1 x
Thus the coefficient of term involving x n is 2n2 + 2n + 1 .
+
(
)
(
)
Since we know that
(1 + x)−1 = 1 − x + x 2 − x3 + ...
Therefore
(v)
(1 − x + x
2
2
2
− x 3 + ...) = ( (1 + x ) −1 ) = (1 + x )
−2
(−2)(−2 − 1) 2 (−2)(−2 − 1)(−2 − 2) 3
( x) +
( x) + ...
2!
3!
(−2)(−3) 2 (−2)(−3)(−4) 3
= 1 − 2x +
( x) +
( x) + ...
2
3⋅ 2
= 1 − 2 x + 3 x 2 − 4 x3 + ...
= 1 + (−1) 2 x + (−1) 2 3x 2 (−1)3 4 x3 + ...
Following in this way we can write
= 1 + (−1) 2 x + (−1) 2 3 x 2 (−1)3 4 x3 + ... + (−1)n (n + 1) x n + ...
= 1 + (−2)( x) +
So the term involving x n = (−1)n (n + 1) x n
And hence coefficient of term involving x n is (−1) n (n + 1)
Question # 4
If x so small that its square and higher powers can be neglected, then show that
1− x
3
1 + 2x
3
≈1+ x
≈= 1 − x
(i)
(ii)
2
2
1+ x
1− x
1
1
(9 + 7 x) 2 − (16 + 3 x) 4 1 17
≈ −
x
(iii)
4 + 5x
4 384
(iv)
4+ x
25
≈2+ x
3
(1 − x)
4
www.mathcity.org
FSc-I / Ex 8.3 - 9
1
(v)
1
(1 + x) 2 (4 − 3x) 4
(8 + 5 x )
1
3
1
 5x 
≈ 1 − 
6 

(vi)
1
(1 − x) 2 (9 − 4 x) 2
(8 + 3x )
1
3
≈
3 61
− x
2 48
1
(vii)
4 − x + (8 − x) 3
1
≈2−
(8 − x )3
1
x
12
Solution
(i)
1
−
1− x
1− x
2
L.H.S =
=
1 = (1 − x )(1 + x )
1 + x (1 + x ) 2
  1

= (1 − x) 1 +  −  ( x) + squares and higher power of x  .
  2

1
= 1 − x − x + squares and higher power of x
2
3
≈ 1 − x = R.H.S Proved
2
(ii)
1
1
1 + 2x
−
= (1 + 2 x ) 2 (1 − x ) 2
1− x
1
1
Now (1 + 2 x ) 2 = 1 +   (2 x) + squares and higher power of x.
2
≈1+ x
1
−
 1
Now (1 − x ) 2 = 1 +  −  (− x) + squares and higher power of x.
 2
1
≈1+ x
2
1 + 2x
 1 
≈ (1 + x ) 1 + x 
1− x
 2 
1
=1+ x + x
ignoring term involving x 2 .
2
3
= 1 + x Proved.
2
Since
1
(iii)
1
1
1
(9 + 7 x) 2 − (16 + 3 x) 4
= (9 + 7 x) 2 − (16 + 3 x) 4
4 + 5x
(
) ( 4 + 5x )
−1
1
 7x 2
Now (9 + 7 x) = 9 1 + 
9 

1
2
1
2
1

 1  7 x 
= (32 ) 2 1 +    + squres and higher of x. 
  2  9 

www.mathcity.org
FSc-I / Ex 8.3 - 10
7x
 7x 
 7x 
≈ 31 +
 = 3 + 3  = 3 +
6
 18 
 18 
1
 3x  4
= (16) 1 + 
 16 
1

 1  3x 
= (24 ) 4 1 +    + square and higher power of x 
  4  16 

3x
 3x 
 3x 
≈ (2) 1 +  = 2 + 2   = 2 +
32
 64 
 64 
1
(16 + 3 x) 4
1
4
−1
( 4 + 5x )
−1
 5 
= 4 1 + x 
 4 
1

5 
= 1 + ( −1)  x  + squares and higher power of x 
4
4 

1 5 
1 5
≈ 1 − x  = − x
4 4 
4 16
−1
1
So
1
(9 + 7 x) 2 − (16 + 3 x) 4  7 x  
3x    1 5 
≈  3 +  −  2 +    − x 
4 + 5x
6  
32    4 16 

3 x   1 5   103  1 5 
 7x
= 3 +
− 2 −   − x  =  1+
x  − x 
6
32   4 16  
96  4 16 

1 103
5
1 17
= +
x− x = −
x Proved
4 384
16
4 384
Do yourself
(iv)
3
1
(v)
(1 + x ) 2 ( 4 − 3x ) 2 = 1 + x 12 4 − 3x 32 8 + 5 x − 13
( ) (
) (
)
1
3
8
+
5
x
(
)
1
1
Now (1 + x ) 2 = 1 +   ( x) + square and higher power of x
2
1
≈1+ x
2
3
( 4 − 3x )
(8 + 5x )
3
2
 3 2
= 4 1 − x 
 4 
3

 3  3 
2 2
= ( 2 ) 1 +   − x  + square and higher power of x 
  2  4 

9 
3
 9 
≈ ( 2 ) 1 − x  = 8  1 − x 
 8 
 8 
−1
3
3
2
= ( 8)
−1
3
 5 
1 + x 
 8 
−1
3
www.mathcity.org
FSc-I / Ex 8.3 - 11
−1 

 1  5 
= ( 23 ) 3 1 +  −  x  + square and higher power of x 
  3  8 

5 
1
5 

≈ (2) −1 1 − x  = 1 − x 
2  24 
 24 
3
1
So
(1 + x ) 2 ( 4 − 3x ) 2
1
8
+
5
x
(
)3
 1   9 
≈ 1 + x  8 1 − x 
 2   8 
1
5 
−
x
1

2  24 
8 1   9
5 
= 1 + x  1 − x − x 
2 2   8
24 
4 
 1  4 
 1
 5 
= 4 1 + x  1 − x  = 4 1 + x − x  = 4 1 − x  Proved
3 
 2  3 
 2
 6 
Do yourself as above
Same as Question #4 (iii)
(vi)
(vii)
Question # 5
If x is so small that its cube and higher power can be neglected, then show that
1+ x
1
1
9
= 1 + x + x2
(i) 1 − x − 2 x 2 = 1 − x − x 2
(ii)
1− x
2
2
8
Solution
1
(i) 1 − x − 2 x 2 = (1 − ( x + 2 x 2 ) ) 2
(
)
1 1 −1
2
1
2
= 1 +   ( −( x + 2 x ) ) + 2 2
−( x + 2 x 2 ) ) + cube & higher power of x.
(
2!
2
1 −1
1
2
≈ 1 −   ( x + 2 x ) + 2 2 ( x + 2 x2 )2
2
2
1
1
1
1
1
≈ 1 − x − (2 x 2 ) − x 2 = 1 − x − x 2 − x 2
2
2
8
2
8
1
9
= 1 − x − x2
Proved
2
8
( )
(ii)
1
x) 2
1
1
−
1 + x (1 +
2
2
=
1 = (1 + x) (1 − x)
1 − x (1 − x) 2
Now
(
)
1 1 −1
1
(1 + x) = 1 +   x + 2 2
x 2 + cube & higher power of x.
2!
2
1 −1
1
1
1
≈ 1 + x + 2 2 x2 = 1 + x − x2
2
2
2
8
1
2
( )
www.mathcity.org
FSc-I / Ex 8.3 - 12
(1 − x)
−
1
2
( )(
)
− 12 − 12 − 1
 1
= 1 +  −  (− x) +
(− x)2 + cube & higher power of x.
2!
 2
− 12 − 23 2
1
1
3
≈1+ x +
x = 1 + x + x2
2
2
2
8
( )( )
So
1+ x  1
1  1
3 
= 1 + x − x 2 1 + x + x 2 
1− x  2
8  2
8 
1
1
1
1
1
3
= 1 + x − x2 + x + x 2 + x2 = 1 + x + x2
2
2
8
2
4
8
Proved
Question # 6
If x is very nearly equal 1, then prove that px p − qx q = ( p − q ) x p + q
Solution
Since x is nearly equal to 1 so suppose x = 1 + h ,
where h is so small that its square and higher powers be neglected
L.H.S = px p − qx q
= p (1 + h) p − q (1 + h) q
= p (1 + ph + square & higher power of x)
−q (1 + qh + square & higher power of h)
≈ p (1 + ph) − q (1 + qh)
= p + p 2 h − q − q 2 h …………….. (i)
Now R.H.S = ( p − q ) x p + q
= ( p − q )(1 + h) p + q
= ( p − q ) (1 + ( p + q )h + square & higher power of h )
≈ ( p − q ) (1 + ( p + q )h ) = ( p − q ) (1 + ph + qh )
= p + p 2 h + pqh − q − pqh − q 2 h
= p + p 2 h − q − q 2 h …………….. (ii)
From (i) and (ii)
L.H.S ≈ R.H.S Proved
Question # 7
If p − q is small when compared with p or q , show that
1
(2n + 1) p + (2n − 1)q  p + q  n
=
 .
(2n − 1) p + (2n + 1)q  2q 
Since p − q is small when compare
Solution
Therefore let p − q = h  p = q + h
(2n + 1) p + (2n − 1)q
(2n + 1)(q + h) + (2n − 1)q
=
L.H.S =
(2n − 1) p + (2n + 1)q
(2n − 1)(q + h) + (2n + 1)q
www.mathcity.org
FSc-I / Ex 8.3 - 13
=
2nq + q + 2nh + h + 2nq − q
2nq − q + 2nh − h + 2nq + q
=
4nq + 2nh + h
4nq + 2nh − h
4nq + 2nh + h
4nq + 2nh + h  2nh − h 
=
=
1 +

 2nh − h 
4nq
4nq 

4nq 1 +
4nq 

−1

 2nh − h 
4nq + 2nh + h 
+ square & higher power of x 2 
1 + (−1) 

4nq
 4nq 


4nq + 2nh + h  2nh − h 
4nq + 2nh + h  4nq − 2nh + h 
=
1 −
 =


4nq
4nq 
4nq
4nq



=
16n 2 q 2 + 8n 2 hq + 4nhq − 8n 2 hq + 4nhq
≈
16n 2 q 2
16n 2 q 2 + 8nhq
16n 2 q 2 8nhq
=
=
+
16n 2 q 2
16n 2 q 2 16n 2 q 2
h
=1+
…………….. (i)
2nq
1
1
1
1
ignoring squares of h
 p + q n  q + h + q n
Now R.H.S = 
 =

2
q

  2q 
1
 2q + h  n  2q h  n 
h n
=
+
 =
 = 1 +

 2q 
 2q 2 q 
 2q 
 1  h 
= 1 +    + square & higher power of h .
 n   2q 
h
≈1+
…………….. (ii)
2nq
Form (i) and (ii)
L.H.S ≈ R.H.S
Proved
Question # 8
1
2


n
8n
n+ N
≈
−
Show that 
where n and N are nearly equal.

9n − N
4n
 2(n + N ) 
Solution Since n and N are nearly equal therefore consider N = n + h ,
where h is so small that its squares and higher power be neglected.
1
1

2 
2
n
n
=
L.H.S = 
 

 2(n + N )   2(n + n + h) 
1
2

  2(2n + h) 
n
=

 =
2(2
n
+
h
)
n


 
−
1
2
 4n + 2h 
=

 n 
−
1
2
2h 

=4+ 
n 

−
1
2
www.mathcity.org
FSc-I / Ex 8.3 - 14
−
1
2
1
2
−
1
2
1
2
h 
 2h 

= (4) 1 +  = (22 )  1 +

 4n 
 2n 
  1 h

= (2) −1 1 +  −  + square & higher power of h 
  2  2n

1
h 
1 h
= 1 −  = −
…………….. (i)
2  4n 
2 8n
8n
n+ N
Now R.H.S =
−
9n − N
4n
8n
n + (n + h)
8n
n+n+h
=
−
=
−
9n − ( n + h)
4n
9n − n − h
4n
−
−
−1
8n
2n + h
h 
2n + h
8n
2n + h 
=
−
= 1 −  −
=
−
4n
8n − h
4n
4n
 8n 
8n 1 − h
8n

  2n h 
 h 
= 1 + (−1)  −  + square & higher power of h  − 
+ 
8
n
4
n
4n 





h  1 h 
h 1 h

= 1 +  −  +
− −
 =1+
8
n
2
4
n
8
n
2 4n

 

1 h
= −
…………….. (ii)
2 8n
From (i) and (ii)
L.H.S = R.H.S
Proved
Question # 9
Identify the following series as binomial expansion and find the sum in each case.
2
3
1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1 
(i) 1 −   +
  −
  +…
2  4  2! ⋅ 4  4 
3! ⋅ 8  4 
(
)
2
3
2
3
1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1 
(ii) 1 −   +
  −
  +…
2  2  2⋅ 4  2 
2.4.6  2 
3 3⋅5 3⋅5⋅ 7
(iii) 1 + +
+
+…
4 4 ⋅8 4 ⋅ 8 ⋅ 12
1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1 
(iv) 1 −   +
  −
  +…
2  3  2⋅4  3 
2.4.6  3 
Solution
2
3
1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1 
(i)
1−   +
  −
  +…
2  4  2! ⋅ 4  4 
3! ⋅ 8  4 
Suppose the given series be identical with
n(n − 1) 2
(1 + x)n = 1 + nx +
x +…
2!
11
This implies
nx = −   ……………….…… (i)
2 4
www.mathcity.org
FSc-I / Ex 8.3 - 15
2
n(n − 1) 2 1 ⋅ 3  1 
x =
  …………. (ii)
2!
2! ⋅ 4  4 
nx = −
From (i)
1
1
 x=−
…………… (iii)
8
8n
Putting value of x in (ii)
2
2
n(n − 1)  1 
1⋅ 3  1 
−  =
 
2!  8n  2! ⋅ 4  4 
n(n − 1)  1 
3 1

=


 
2  64n 2  2 ⋅ 4  16 
(n − 1)
3
3

=
 (n − 1) =
⋅ 128n  n − 1 = 3 n
128n 128
128
1
 n − 3n = 1  − 2n = 1  n = −
2
Putting value of n in equation (iii)
1
1
 x=
x=−
4
8 −1
2
So
( )
−
(ii)
1
 1 2 5
(1 + x) =  1 +  =  
 4
4
Do yourself as above
n
−
1
2
1
4
 4 2
=  =
5
5
3 3⋅5 3⋅5⋅ 7
+
+
+…
4 4 ⋅8 4 ⋅ 8 ⋅ 12
Suppose the given series be identical with
n(n − 1) 2
(1 + x)n = 1 + nx +
x +…
2!
3
nx = ……………….…… (i)
This implies
4
n(n − 1) 2 3 ⋅ 5
x =
…………. (ii)
2!
4 ⋅8
3
3
nx =
From (i)
 x=
…………… (iii)
4
4n
Putting value of x in (ii)
2
n(n − 1)  3  3 ⋅ 5
n(n − 1)  9  15
=


=
 
2!  4n  4 ⋅8
2  16 n 2  32
9(n − 1) 15
15

=
 9( n − 1) = ⋅ 32 n  9n − 9 = 15 n
32 n
32
32
(iii)
1+
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FSc-I / Ex 8.3 - 16
 9n − 15n = 9  − 6n = 9  n = −
9
6
 n=−
3
2
Putting value of n in equation (iii)
1
3
 x=−
x=−
2
4 −3
2
( )
−
3
2
−
3
2
3
3
 1
1
So (1 + x) =  1 −  =   = ( 2 ) 2 = 2 = 2 2 Answer
 2
2
(iv)
Do yourself as above
Question # 10
1 1⋅ 3 1⋅ 3 ⋅ 5
Use binomial theorem to show that 1 + +
+
+… = 2
4 4 ⋅8 4 ⋅ 8 ⋅ 12
1 1⋅ 3 1⋅ 3 ⋅ 5
Solution
1+ +
+
+…
4 4 ⋅8 4 ⋅ 8 ⋅ 12
Suppose the given series be identical with
n(n − 1) 2
(1 + x)n = 1 + nx +
x +…
2!
This implies
1
nx = ……………….…… (i)
4
n(n − 1) 2 1 ⋅ 3
x =
…………. (ii)
2!
4 ⋅8
1
1
nx =
From (i)
 x=
…………… (iii)
4
4n
Putting value of x in (ii)
2
n( n − 1)  1  1 ⋅ 3
n( n − 1)  1  3
=


=
 
2!  4n  4 ⋅8
2  16 n 2  32
(n − 1) 3
3

=
 (n − 1) = ⋅ 32 n  n − 1 = 3 n
32 n
32
32
n
( )
 n − 3n = 1  − 2n = 1
Putting value of n in equation (iii)
1

x=
4 −1
2
x=−
( )
−
So
Hence
1
−
 n=−
1
2
1
2
1
1
 1 2 1 2
(1 + x) n =  1 −  =   = ( 2 ) 2 = 2
 2
2
1 1⋅ 3 1⋅ 3 ⋅ 5
1+ +
+
+… = 2
Proved
4 4 ⋅8 4 ⋅ 8 ⋅ 12
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FSc-I / Ex 8.3 - 17
Question # 11
2
3
1 1⋅ 3  1  1⋅ 3 ⋅ 5  1 
2
If y = +
  +
  + … , then prove that y + 2 y − 2 = 0 .
3 2!  3 
3!  3 
2
3
1 1⋅ 3  1  1⋅ 3 ⋅ 5  1 
Solution
y= +
  +
  +…
3 2!  3 
3!  3 
Adding 1 on both sides
2
3
1 1⋅ 3  1  1⋅ 3 ⋅ 5  1 
1+ y =1+ +
  +
  +…
3 2!  3 
3!  3 
Let the given series be identical with
n(n − 1) 2
(1 + x)n = 1 + nx +
x +…
2!
This implies
1
nx = ……………….…… (i)
3
2
n(n − 1) 2 1 ⋅ 3  1 
x =
  …………. (ii)
2!
2!  3 
1
1
From (i)
nx =
 x=
…………… (iii)
3
3n
Putting value of x in (ii)
2
2
n(n − 1)  1  1 ⋅ 3  1 
  =
 
2!  3n 
2!  3 
n( n − 1)  1  3 1


= ⋅
2  9 n2  2 9
n −1 1
1

=
 n − 1 = ⋅ 18 n
18 n 6
6
 n − 1 = 3 n  n − 3n = 1
 − 2n = 1  n = −
1
2
Putting value of n in equation (iii)
2
1
 x=−
x=
3
3 −1
2
( )
 2
So (1 + x) n =  1 − 
 3
−
1
2
1
= 
3
−
1
2
1
2
= ( 3) = 3
This implies
1+ y = 3
On squaring both sides
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FSc-I / Ex 8.3 - 18
(1 + y )
2
2
=
( )
3
 1 + 2 y + y2 = 3  1 + 2 y + y2 − 3 = 0
 y 2 + 2 y − 2 = 0 Proved.
Question # 12
1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
If 2 y = 2 +
⋅ 4+
⋅ 6 + ... , then prove that 4 y 2 + 4 y − 1 = 0 .
2
2! 2
3! 2
1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
Solution
2y = 2 +
⋅ +
⋅ + ...
2
2! 24
3! 26
Adding 1 on both sides
1 1⋅ 3 1 1⋅ 3 ⋅ 5 1
1+ 2y =1+ 2 +
⋅ +
⋅ + ...
2
2! 24
3! 26
Let the given series be identical with
n(n − 1) 2
x + ...
(1 + x)n = 1 + nx +
2!
This implies
1
nx = 2 ……………….…… (i)
2
n(n − 1) 2 1 ⋅ 3 1
x =
⋅ …………. (ii)
2!
2! 24
1
1
nx =
From (i)
 x=
…………… (iii)
4
4n
Putting value of x in (ii)
2
n(n − 1)  1  1 ⋅ 3 1
⋅
  =
2!  4n 
2! 24
n( n − 1)  1  3 1


= ⋅
2  16 n 2  2 16
n −1

= 3  n − 1 = 3n
n
1
 n − 3n = 1  − 2n = 1  n = −
2
Putting value of n in equation (iii)
1
1
 x=−
x=
2
4 −1
2
So
( )
 1
(1 + x) n =  1 − 
 2
1
= 
2
−
1
2
−
1
2
= ( 2)
1
2
= 2
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FSc-I / Ex 8.3 - 19
This implies
1+ 2y = 2
On squaring both sides
(1 + 2 y )
2
=
( 2)
 1 + 4 y + 4 y2 = 4
 4 y2 + 4 y − 1 = 0
2
 1 + 4 y + 4 y2 − 2 = 0
Proved
Question # 13
2
3
2 1⋅ 3  2  1⋅ 3 ⋅ 5  2 
2
If y = +
  +
  + … , then prove that y + 2 y − 4 = 0 .
5 2!  5 
3!  5 
2
3
2 1⋅ 3  2  1⋅ 3 ⋅ 5  2 
Solution
y= +
  +
  +…
5 2!  5 
3!  5 
Adding 1 on both sides
2
3
2 1⋅ 3  2  1⋅ 3 ⋅ 5  2 
1+ y =1+ +
  +
  +…
5 2!  5 
3!  5 
Let the given series be identical with
n(n − 1) 2
(1 + x)n = 1 + nx +
x +…
2!
This implies
2
nx = ……………….…… (i)
5
2
n( n − 1) 2 1 ⋅ 3  2 
x =
  …………. (ii)
2!
2!  5 
2
2
From (i)
nx =
 x=
…………… (iii)
5
5n
Putting value of x in (ii)
2
2
n(n − 1)  2  1 ⋅ 3  2 
  =
 
2!  5n 
2!  5 
n(n − 1)  4  3  4 


=  
2  25n 2  2  25 
n −1

= 3  n − 1 = 3 n  n − 3n = 1
n
1
 − 2n = 1  n = −
2
Putting value of n in equation (iii)
4
2
 x=−
x=
5
5 −1
2
( )
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FSc-I / Ex 8.3 - 20
 4
So (1 + x) n =  1 − 
 5
−
1
2
1
= 
5
−
1
2
1
= ( 5) 2 = 5
This implies
1+ y = 5
On squaring both sides
(1 + y )
2
=
( 5)
2
 1 + 2 y + y2 = 5  1 + 2 y + y2 − 5 = 0
 y 2 + 2 y − 4 = 0 Proved.
If you found any error, please report us at www.mathcity.org/error
Error Analyst
Faizan Yousaf
Book:
(2019)
Punjab College
Exercise 8.3 (Page 283)
Text Book of Algebra and Trigonometry Class XI
Punjab Textbook Board, Lahore.
Edition: May 2017
Available online at http://www.MathCity.org in PDF Format
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Updated: March 16,2019.
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