HOW TO FIND EXPANSION USING BINOMIAL THEOREM

Binomial expansion for (x + a)n is,

nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + .........+ ncnxn-na0

Question 1 :

Expand (i) [2x2 − (3/x)]3

Solution :

x  =  2x2,  a  =  (-3/x),  n  =  3

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  3c0 (2x2)3(-3/x)0 + 3c1 (2x2)2(-3/x)1 + 3c2 (2x2)1(-3/x)2 + 3c3 (2x2)0(-3/x)3

3c0  =  1

3c1  =  3

3c2  =  3

3c3  =  1

  =  (8x6) + 3(4x4)(-3/x) + 3(2x2)(9/x2) + 1 (1)(-27/x3)

  =  8x6 - 36x3 + 54 - (27/x3)

Question 2 :

Expand (ii)  (2x2 − 3√1 − x2)4  +  (2x2 + 3√1 − x2)4

Solution :

Part 1 :

  =  (2x2 − 3√1 − x2)4  

x  =  2x2,  a  =  (-3√1 − x2),  n  =  4

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  4c0 (2x2)4(-3√1 −x2)0

+ 4c1 (2x2)3(-3√1−x2)1 + 4c2 (2x2)2(-3√1−x2)2

+ 4c3 (2x2)1(-3√1 −x2)3+ 4c4 (2x2)0(-3√1 −x2)4

4c0  =  1

4c1  =  4

4c2  =  6

4c3  =  4

4c4  =  1

=  16x8

+ 4 (8x6)(-3√1−x2)1 + 6 (4x4) (9√1−x2)2

+ 4c3 (2x2)1(-27√1 −x2)3+ 4c4 (2x2)0(-3√1 −x2)4

  =  16x8

- 96x6 √1−x2 + 216x4 √1−x2 - 216x2(1-x2) √1−x2 

  + 81 (1 −x2)2  --------(1)

Part 2 :

  =  (2x2 + 3√1 − x2)4  

x  =  2x2,  a  =  (3√1 − x2),  n  =  4

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  4c0 (2x2)4(3√1 −x2)0

+ 4c1 (2x2)3(3√1−x2)1 + 4c2 (2x2)2(3√1−x2)2

+ 4c3 (2x2)1(3√1 −x2)3+ 4c4 (2x2)0(3√1 −x2)4

  =  16x8

+ 96x6 √1−x2 + 216x4 √1−x2 + 216x2(1-x2) √1−x2 

  + 81 (1 −x2)2  --------(2)

(1) + (2)  ==>  2 [16x8 + 216x4 √1−x2 + 81 (1 −x2)2]

Hence the answer is 2 [16x8 + 216x4 √1−x2 + 81 (1 −x2)2].

Question 3 :

Compute (i)  1024

Solution :

1024  =  (100 + 2)4

x  =  100, a  =  2, n  =  4

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  4c0 (100)4(2)0 + 4c1(100)3(2)1 + 4c2(100)2(2)2 + 4c3(100)1(2)3 + 4c4(100)0(2)4

4c0  =  1

4c1  =  4

4c2  =  6

4c3  =  4

4c4  =  1

  =  1 (100000000)(1) + 4(1000000)(2) + 6(10000)(4) + 4(100)1(8) + 1(1)(16)

  =  100000000 + 8000000 + 240000 + 3200 + 16

  =  108243216

Hence the value of  1024 is 108243216.

Question 4 :

Compute (ii)  994

Solution :

994  =  (100 - 1)4

x  =  100, a  =  -1, n  =  4

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  4c0 (100)4(-1)0 + 4c1(100)3(-1)1 + 4c2(100)2(-1)2 + 4c3(100)1(-1)3 + 4c4(100)0(-1)4

4c0  =  1

4c1  =  4

4c2  =  6

4c3  =  4

4c4  =  1

  =  1 (100000000)(1) + 4(1000000)(-1) + 6(10000)(1) + 4(100)1(-1) + 1(1)(1)

  =  100000000 - 4000000 + 60000 - 400 + 1

  =  96059601

Hence the value of  994 is 96059601.

Question 5 :

Compute (iii)  97

Solution :

97  =  (10 - 1)7

x  =  10, a  =  -1, n  =  7

= nc0xna0 + nc1xn-1a1 + nc2xn-2a2 + ........+ ncnxn-na0

=  7c0 (10)7(-1)0 + 7c1(10)6(-1)1 + 7c2(10)5(-1)2 + 7c3(10)4(-1)3 + 7c4(10)3(-1)4+ 7c5(10)2(-1)5+ 7c6(10)0(-1)6  + 7c7(10)1(-1)7

  =  1(10000000) - 7(1000000) + 21(100000) - 35(10000) + 35(1000) - 21(100) + 7 - 1

  =  10000000-7000000+2100000-350000+35000- 2100+70-1

=  4782969 

Hence the value of  97 is 4782969.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

RSS

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Dec 23, 24 03:47 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 91)

    Dec 23, 24 03:40 AM

    Digital SAT Math Problems and Solutions (Part - 91)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 90)

    Dec 21, 24 02:19 AM

    Digital SAT Math Problems and Solutions (Part - 90)

    Read More