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## The Binomial Theorem

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

What do you notice in the above expansion of various powers of (x + y)?

1. The number of terms in the expansion is one more than the exponent.

2. In each expansion:
1. The exponent of the first term is same as the exponent of the binomial. The exponent of y in the first term is zero.

2. Subsequently, in each successive term, the exponent of x decreases by 1 with a simultaneous increase of 1 in the exponent of y.

3. The sum of the exponents of x and y in each term is equal to the exponent of the binomial.

4. The exponent of x in the last term is zero and that of y is equal to the exponent of the binomial.

### Binomial theorem for a positive integral exponent

Theorem:

If n is a natural number,

### Example3

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.

### Try these questions

 1. Expand (x + 1/y)7 Find the middle term of the following expansion 2. (x/a + y/b)6 3. (√a – b)8 4. (x2/y – y2/x)8 5. (xy – 1/x2y2)5 6. (a/x + x/a)5 7. (x – 3/y)5 8. Find the term containing x5 in the expansion of (x – 1/x)11 9. Write the 14th term in the expansion of (3 + x)15 10. Write the 10th term in the expansion of (3 + x)12