Binomial Theorem 2
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The document discusses the application of the binomial theorem to find coefficients of specific terms in a polynomial expression. It outlines several calculation steps to derive values for variables a and b under certain conditions. The final result reveals the relationship between these variables based on the coefficients extracted from the expansion.
![Binomial Theorem 2
Physics Helpline
L K Satapathy
Answer :
2 18
(1 )(1 2 )ax bx x
2
0 1 2(1 ) . . ... ( 1) . .n n n n n n n
nx C C x C x C x Binomial Theorem :
Step-1 The given expression
2 18 18 2 18 3 18 4
1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x
Using Binomial Theorem , we get
2 18
(1 )(1 2 )ax bx x
18 18 18 2 18 3 18 4
1 2 3 4(1 2 ) [1 (2 ) (2 ) (2 ) (2 ) . . . ]x C x C x C x C x ](https://image.slidesharecdn.com/3-160112125453/85/Binomial-Theorem-2-3-320.jpg)
![Binomial Theorem 2
Physics Helpline
L K Satapathy
18 18 18
3 2 18 4 2 0C a C b C
Step-2
51 3 544 0 . . . (1)a b
3 18 3 18 2 2 18
3 2 1: .(2 ) . .(2 ) . (2 )x termsare C x ax C x bx C x
18.17.16 18.178 4 2 18 0
3.2 2
a b
2 18 18 2 18 3 18 4
1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x
Pairs of terms producing :3
x
Putting the coefficient of , we get3
0x ](https://image.slidesharecdn.com/3-160112125453/85/Binomial-Theorem-2-4-320.jpg)
![Binomial Theorem 2
Physics Helpline
L K Satapathy
Step-3
4 18 4 18 3 2 18 2
4 3 2: .(2 ) . .(2 ) . (2 )x termsare C x ax C x bx C x
32 3 240 0 . . . (2)a b
18 18 18
4 3 216 8 4 0C a C b C
18.17.16.15 18.17.16 18.1716 8 4 0
4.3.2 3.2 2
a b
2 18 18 2 18 3 18 4
1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x
Pairs of terms producing :4
x
Putting the coefficient of , we get
4
0x ](https://image.slidesharecdn.com/3-160112125453/85/Binomial-Theorem-2-5-320.jpg)
Binomial Theorem 2
- 1. Physics Helpline L K Satapathy Binomial Theorem 2
- 2. Binomial Theorem 2 Physics Helpline L K Satapathy 251 251 272 272( ) 16, ( ) 14, ( ) 14, ( ) 16, 3 3 3 3 a b c d Q : If the coefficients of and in the expansion of in powers of x are both zero , then (a , b) is equal to 3 x 4 x 2 18 (1 )(1 2 )ax bx x
- 3. Binomial Theorem 2 Physics Helpline L K Satapathy Answer : 2 18 (1 )(1 2 )ax bx x 2 0 1 2(1 ) . . ... ( 1) . .n n n n n n n nx C C x C x C x Binomial Theorem : Step-1 The given expression 2 18 18 2 18 3 18 4 1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x Using Binomial Theorem , we get 2 18 (1 )(1 2 )ax bx x 18 18 18 2 18 3 18 4 1 2 3 4(1 2 ) [1 (2 ) (2 ) (2 ) (2 ) . . . ]x C x C x C x C x
- 4. Binomial Theorem 2 Physics Helpline L K Satapathy 18 18 18 3 2 18 4 2 0C a C b C Step-2 51 3 544 0 . . . (1)a b 3 18 3 18 2 2 18 3 2 1: .(2 ) . .(2 ) . (2 )x termsare C x ax C x bx C x 18.17.16 18.178 4 2 18 0 3.2 2 a b 2 18 18 2 18 3 18 4 1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x Pairs of terms producing :3 x Putting the coefficient of , we get3 0x
- 5. Binomial Theorem 2 Physics Helpline L K Satapathy Step-3 4 18 4 18 3 2 18 2 4 3 2: .(2 ) . .(2 ) . (2 )x termsare C x ax C x bx C x 32 3 240 0 . . . (2)a b 18 18 18 4 3 216 8 4 0C a C b C 18.17.16.15 18.17.16 18.1716 8 4 0 4.3.2 3.2 2 a b 2 18 18 2 18 3 18 4 1 2 3 4(1 )[1 (2 ) (2 ) (2 ) (2 ) . . . ]ax bx C x C x C x C x Pairs of terms producing :4 x Putting the coefficient of , we get 4 0x
- 6. Binomial Theorem 2 Physics Helpline L K Satapathy Correct option = (d) (1) (2) 51 32 544 240 304a a (2) 32 3 240a b Step-4 (2) 3 2 240 32 16 240 272b a (1) 51 3 544a b 19 304 16a a 2723 272 3 b b 272( , ) 16, 3 a b
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