Ex-8-3-fsc-part1-ver3-5.pdf

MathCity.org Merging man and maths

Exercise 8.3 (Solutions) Textbook of Algebra and Trigonometry for Class XI Available online @ http://www.mathcity.org, Version: 3.5

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 x + x + ... (1 + x) n = 1 + n 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 − 3 x)

−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 + (− x 3 ) + ... 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 − 1 − 2 2 3       1  3x  2 2    − 3 x  + 2  2  2   − 3x  + ... = 2 1 +  −  +      2!  4  3!  4   2 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   3 x 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)   (x ) + (− x 3 ) + …  = 1 + x + 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 x 3 +…

(

) (

)

= 1 + ( x − 2 x) + ( x 2 − 2 x 2 + 3x 2 ) + ( x3 − 2 x3 + 3x3 − 4 x3 ) + … = 1 − x + 2 x 2 − 2 x 3 +…

Do yourself as above

(viii) 1

1

1 1 (4 + 2 x) 2 2x 2  x = (4 + 2 x) 2 (2 − x) −1 = (4) 2 1 +  (2) −1 1 −  (ix) 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

www.mathcity.org

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     −   x 3   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 2 4 3 ⋅ 2 8           x x2 x3 x x 2 x3 = 1 + − + + …  × 1 + + + + …  8  4 32 128   2 4  2 x2 x2   x3 x3 x3 x3  x x  x =1+  +  + − + +  +  − + +  +… 4 2 32 8 4 128 64 16 8       3 x 11x 2 23 x3 =1+ + + +… 4 32 128 1

1

(1 + x − 2 x 2 ) 2 = (1 + ( x − 2 x2 ) ) 2

(x)

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 x3 + 3( x) 2 (−2 x 2 ) + 3( x)(−2 x 2 ) 2 − (2 x 2 )3 +…

(

(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 + x 3 − x 4 + x5 − x 6 + … 2 2 8 8 8 16 16 16 16 1 1 1 1 1 3 3 1 = 1 + x − x2 − x2 − x3 + x4 + x3 − x 4 + x5 − x6 + … 2 8 2 2 16 8 4 8 1 9 9 = 1 + x − x2 − x3 + … 2 8 16 Do yourself as above www.mathcity.org

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

65 1 (viii) 5 252 1

(xii) (1280) 4

Solution (i)

1

(ii) (iii)

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 + …) ≈ 10 ( 0.9949875 ) = 9.949875 ≈ 9.950 1 2

1 2

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

www.mathcity.org

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  3 3   − 1  + … =   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 www.mathcity.org

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 + x2 ) ( 1 + x) ( (ii) (i) 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 (i) = (1 + x 2 ) (1 + x ) 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 x 3 + …) = (1 + x 2 )(1 + (−1)2 x + (−1) 2 3x 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) xn−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 = ( n + 1 + n − 1)( −1) n x n = (2n)( −1) n x n

∵ ( −1) −2 = 1

www.mathcity.org

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

(1 + x ) = 2

(1 − x )

2

(1 + x ) (1 + 2 x + 3x 2

2

+ 4 x 3 + … + (n − 1) x n−2 + (n) x n−1 + (n + 1) x n + …) 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) + (− x3 ) + …  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 + x 3 (1 + 2 x + 3x 2 + 4 x 3 + … + (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 + 2 n + 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 2n 2 + 2n + 1 .

+

(

(v)

)

(

)

Since we know that (1 + x )−1 = 1 − x + x 2 − x 3 + ............... Therefore

(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 x 3 + ................... = 1 + (−1) 2 x + (−1) 2 3 x 2 (−1)3 4 x 3 + ................... Following in this way we can write = 1 + (−1) 2 x + (−1) 2 3 x 2 (−1)3 4 x 3 + ................... + (−1)n (n + 1) x n + ......... 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) = 1 + (−2)( x) +

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 (i) ≈= 1 − x (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 (1 − x)3 4 www.mathcity.org

FSc-I / Ex 8.3 - 9 1

(v)

1

(8 + 5x ) (vii)

1

 5x  ≈ 1 −  6  

(1 + x) 2 (4 − 3 x) 4 1 3

4 − x + (8 − x)

1 3

1

≈2−

(8 − x )3

(vi)

1

(1 − x) 2 (9 − 4 x) 2

(8 + 3x )

1 3

≈

3 61 − x 2 48

1 x 12

Solution (i) 1 1 1− 1− x 1− x 2 = = (1 − x) 2 L.H.S = 1 = (1 − x ) 1 − x (1 − x ) 2

1 = 1 +   (− 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   7x  7x   7x  ≈ 3 1 +  = 3 + 3  = 3 + 6  18   18 

www.mathcity.org

FSc-I / Ex 8.3 - 10 1

 3x  4 = (16) 1 +   16  1    1  3 x  = (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 23 8 + 5 x − 13 ( ) ( ) ( ) 1 3 (8 + 5x ) 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   = ( 22 ) 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

−1

 5  3 = ( 8 ) 1 + x   8  −1    1  5  = ( 23 ) 3 1 +  −  x  + square and higher power of x    3  8   −1 3

www.mathcity.org

FSc-I / Ex 8.3 - 11

5  1 5   ≈ (2)−1 1 − x  = 1 − x  2  24   24  3

1

So

(1 + x ) 2 ( 4 − 3x ) 2 1 (8 + 5x ) 3

5   1   9  1 ≈ 1 + x  8 1 − x  1 − x   2   8  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 9 1+ x 1 = 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 x 2 )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

1 1 − 1 + x (1 + x) 2 2 = = (1 + x) (1 − x) 2 1 − x (1 − x) 12

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 − 12 − 12 − 1 −  1 2 (1 − x) = 1 +  −  (− x) + (− x)2 + cube & higher power of x. 2!  2 1 2

( )

( )(

)

www.mathcity.org

FSc-I / Ex 8.3 - 12

( )( )

− 12 − 23 2 1 3 1 ≈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 3 1 = 1 + x − x2 + x + x2 + 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  Solution Since p − q is small when compare 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 2nq + q + 2nh + h + 2nq − q 4nq + 2nh + h = = 2nq − q + 2nh − h + 2nq + q 4nq + 2nh − h

www.mathcity.org

FSc-I / Ex 8.3 - 13

4nq + 2nh + h 4nq + 2nh + h  2nh − h  = = 1 +   2nh − h  4nq 4nq   4nq 1 + 4nq  

−1

 2nh − h  4nq + 2nh + h  2 & + square higher power of x 1 + (−1)    4nq 4 nq     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 =   =  2q  2 q     1

 2q + h  n  2 q 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 2

    n n L.H.S =   =   2(n + N )   2(n + n + h)  1 2

   2(2n + h)  n =   = n   2(2n + h)   1 2

 2h  = (4) 1 +   4n  −

−

1 2

1 2

−

1 2

 4n + 2h  =   n 

h   = (22 )  1 +   2n  −

1 2

−

−

1 2

2h   = 4+  n  

−

1 2

1 2

www.mathcity.org

FSc-I / Ex 8.3 - 14

  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  −  +   8n     4n 4n  h  1 h  h 1 h  = 1 +  −  +  = 1 + − − 8n 2 4n  8n   2 4 n  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 nx = −   ……………….…… (i) This implies 2 4 2 n(n − 1) 2 1 ⋅ 3  1  x =   …………. (ii) 2! 2! ⋅ 4  4  www.mathcity.org

FSc-I / Ex 8.3 - 15

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  2  64n  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) n =  1 +  =    4 4 Do yourself as above

−

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 15 9(n − 1) 15 ⇒ = ⇒ 9( n − 1) = ⋅ 32 n ⇒ 9n − 9 = 15 n 32 n 32 32

(iii)

1+

⇒ 9n − 15n = 9 ⇒ − 6n = 9 ⇒ n = −

9 6

⇒ n=−

3 2

Putting value of n in equation (iii) www.mathcity.org

FSc-I / Ex 8.3 - 16

x=−

3

( )

4 −3 2 −

⇒

3

x=− −

1 2

3

3 3  1 2 1 2 So (1 + x) n =  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 1+ + Solution + +… 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 3 (n − 1) 3 ⇒ = ⇒ (n − 1) = ⋅ 32 n ⇒ n − 1 = 3 n 32 n 32 32

( )

⇒ n − 3n = 1 ⇒ − 2n = 1

Putting value of n in equation (iii) 1 x= ⇒ 1 4 − 2

x=−

( )

−

So Hence

1 2

−

⇒ n=−

1 2

1 2

1 2

1  1 1 (1 + x) n =  1 −  =   = ( 2 ) 2 = 2  2 2 1 1⋅ 3 1⋅ 3 ⋅ 5 1+ + + +… = 2 Proved 4 4 ⋅8 4 ⋅ 8 ⋅ 12

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  www.mathcity.org

FSc-I / Ex 8.3 - 17 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 1 n −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) 1 2 x= ⇒ x=− 3 3 −1 2

( )

 2 So (1 + x) n =  1 −   3

−

1 2

1 =  3

−

1 2

1

= ( 3) 2 = 3 This implies 1+ y = 3 On squaring both sides

(1 + y )

2

=

( 3)

2

⇒ 1 + 2 y + y2 = 3 ⇒ 1 + 2 y + y2 − 3 = 0 ⇒ y 2 + 2 y − 2 = 0 Proved. www.mathcity.org

FSc-I / Ex 8.3 - 18

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 = ⇒ x= …………… (iii) From (i) 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

This implies 1+ 2y = 2 On squaring both sides

(1 + 2 y )

2

=

( )

2

2

www.mathcity.org

FSc-I / Ex 8.3 - 19

⇒ 1 + 4 y + 4 y2 = 4

⇒ 1 + 4 y + 4 y2 − 2 = 0 Proved

⇒ 4 y2 + 4 y −1 = 0

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  y= + Solution   +   +… 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 nx = ⇒ x= …………… (iii) From (i) 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) 2 4 x= ⇒ x=− 5 5 −1 2

( )

 4 So (1 + x) n =  1 −   5

−

1 2

1 =  5

−

1 2

1

= ( 5) 2 = 5 This implies 1+ y = 5 www.mathcity.org

FSc-I / Ex 8.3 - 20

On squaring both sides 2

(1 + y ) =

( )

2

5

⇒ 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

Book:

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 (Picture format to view online). Page setup: A4 (8.27 in × 11.02 in). Updated: November 13,2017.

These resources are shared under the licence AttributionNonCommercial-NoDerivatives 4.0 International https://creativecommons.org/licenses/by-nc-nd/4.0/ Under this licence if you remix, transform, or build upon the material, you may not distribute the modified material.

www.mathcity.org