HOW TO FIND EXPANSION USING BINOMIAL THEOREM

Binomial expansion for (x + a)n is,

nc0xna+ nc1xn-1a+ nc2xn-2a+ .........+ ncnxn-na0

Question 1 :

Expand (i) [2x2 − (3/x)]3

Solution :

x  =  2x2,  a  =  (-3/x),  n  =  3

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  3c(2x2)3(-3/x)3c(2x2)2(-3/x)3c(2x2)1(-3/x)3c(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 − 31 − x2)+  (2x2 + 31 − x2)4

Solution :

Part 1 :

  =  (2x2 − 31 − x2)4  

x  =  2x2,  a  =  (-31 − x2),  n  =  4

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  4c(2x2)4(-31 −x2)0

4c(2x2)3(-31−x2)4c(2x2)2(-31−x2)2

4c(2x2)1(-31 −x2)34c(2x2)0(-31 −x2)4

4c0  =  1

4c1  =  4

4c2  =  6

4c3  =  4

4c4  =  1

=  16x8

4 (8x6)(-31−x2)+ 6 (4x4) (91−x2)2

4c(2x2)1(-271 −x2)34c(2x2)0(-31 −x2)4

  =  16x8

- 96x1−x2 + 216x1−x- 216x2(1-x21−x

  + 81 (1 −x2) --------(1)

Part 2 :

  =  (2x2 + 31 − x2)4  

x  =  2x2,  a  =  (31 − x2),  n  =  4

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  4c(2x2)4(31 −x2)0

4c(2x2)3(31−x2)4c(2x2)2(31−x2)2

4c(2x2)1(31 −x2)34c(2x2)0(31 −x2)4

  =  16x8

+ 96x1−x2 + 216x1−x+ 216x2(1-x21−x

  + 81 (1 −x2) --------(2)

(1) + (2)  ==>  2 [16x+ 216x1−x+ 81 (1 −x2)2]

Hence the answer is 2 [16x+ 216x1−x+ 81 (1 −x2)2].

Question 3 :

Compute (i)  1024

Solution :

102=  (100 + 2)4

x  =  100, a  =  2, n  =  4

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  4c(100)4(2)+ 4c1(100)3(2)+ 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

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  4c(100)4(-1)+ 4c1(100)3(-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  99is 96059601.

Question 5 :

Compute (iii)  97

Solution :

97  =  (10 - 1)7

x  =  10, a  =  -1, n  =  7

= nc0xna+ nc1xn-1a+ nc2xn-2a+ ........+ ncnxn-na0

=  7c(10)7(-1)+ 7c1(10)6(-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  9is 4782969.

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