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.
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