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We know that any algebraic expression with two variables is called a binomial.
Observe the following powers of the binomial (x + y).
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = (x + y)3(x + y)
= (x3 + 3x2y + 3xy2 + y3) (x + y)
= x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3+ y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 |
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| What do you notice in the above expansion of various powers of (x + y)? |
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- The number of terms in the expansion is one more than the exponent.
- In each expansion:
- The exponent of the first term is same as the
exponent of the binomial. The exponent of y in the first term is
zero.
- Subsequently, in each successive term, the
exponent of x decreases by 1 with a simultaneous increase of 1 in
the exponent of y.
- The sum of the exponents of x and y in each term is equal to the exponent of the binomial.
- The exponent of x in the last term is zero and that of y is equal to the exponent of the binomial.
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Binomial theorem for a positive integral exponent
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| If n is a natural number, |
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Example 1
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Example 2
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Example 3
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Find the constant term or the term independent of x in the expansion of (3x – 5/x2 )9.
Solution:
Tr +1 = 9Cr (3x)9-r * (–5/x9)r
= 9Cr 39-r * x9-r * (–1)r * 5r/x2r
=9Cr 39-r* 5r (–1)r * x9-3r ———————— (i)
To get the constant term of the expansion we have to find r so that 9 – 3r = 0
⇒ r = 3
∴ 4th term is the one independent of x, i.e., the constant term. |
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Try these questions
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1.
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Expand (x + 1/y)7 |
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Find the middle term of the following expansion |
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2.
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(x/a + y/b)6 |
3.
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(√a – b)8 |
4.
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(x2/y – y2/x)8 |
5.
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(xy – 1/x2y2)5 |
6.
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(a/x + x/a)5 |
7.
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(x – 3/y)5 |
8.
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Find the term containing x5 in the expansion of (x – 1/x)11 |
9.
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Write the 14th term in the expansion of (3 + x)15 |
10.
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Write the 10th term in the expansion of (3 + x)12 |
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Answers
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1.
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Expand (x + 1/y)7 |
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Solution: |
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(x + 1/y)7 = 7C0 x7 + 7C1 x6(1/y) + 7C2 x5 (1/y)2 + 7C3 x4 (1/y)3 + 7C4 x3(1/y)4
+ 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + 7C7 (1/y)7
= x7 + 7x6(1/y) + 7C2 x5(1/y)2 + 7C3 x4(1/y)3 + 7C4 x4(1/y)4
+ 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + (1/y))7
here
7! = 7 *6 *5 *4 *3 *2 *1
7! 7! 7 *6 *5 *4 *3 *2 *1
7C2 = ———— = —— = —————————— = 21
(7 – 2)!2! 5!2! 5 *4 *3 *2 *1 *2 *1
7! 7! 7 * 6 * 5
7C3 = ———— = —— = ———— = 35
(7 – 3)!3! 4!3! 3 * 2 *1
7C4 = 7C3 = 35 (since nCr = nCn-r )
7C5 = 7C2 = 21
7C6 = 7C1 = 7
∴ (x + 1/y)7 = x7 + 7x6(1/y) + 21x5(1/y)2 + 35x4(1/y)3 + 35x3(1/y)4 + 21x2(1/y)5
+ 7x(1/y)6 + (1/y)7 |
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2.
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(x/a + y/b)6
Solution:
Expansion contains 7 terms in it and 4th term is the middle term. |
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T4= T1+3 = 6C3 (x/a)6-3 (y/b)3
= 6C3 (x/a)3 (y/b)3 |
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3.
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(√a – b)8
Solution:
Expansion contains 9 terms in it and 5th term is the middle term.
T5 = T1+4 = 8C4 (√a)8-4 (b)4
= 8C4 (√a)4 (b)4
= 8C4 a2b4 |
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4.
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(x2/y – y2/x)8
Solution:
Expansion contains 9 terms in it and 5th term is the middle term.
T5 = T1+4 = 8C4 (x2/y)8–4 (–y2/x)4
= 8C4 (x2/y)4 (–y2/x)4
x8 * y8
= 8C4 ————
x4 * y4
= 8C4 x4 y4 |
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5.
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(xy – 1/x2 y2)5
Solution:
Expansion contains 6 terms in it and 3rd and 4th terms are the middle terms. |
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6.
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(a/x + x/a)5
Solution:
Expansion contains 5 terms in it and 3rd term is the middle term. |
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7.
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(x – 3/y)5
Solution:
Expansion contains 10 terms in it and 5th and 6th terms are the middle terms. |
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8.
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Find the term containing x5 in the expansion of (x – 1/x)11
Solution: |
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9.
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Write the 14th term in the expansion of (3 + x)15
Solution: |
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10.
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Write the 10th term in the expansion of (3 + x)12
Solution: |
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