**Binomial Expression**

Any algebraic expression consisting of only two terms is known as a **binomial expression. **

It’s expansion in power of x is known as the binomial expansion.

e.g. (i) a + x (ii) a^{2} + 1/x^{2} (iii) 4x − 6y

**Binomial Theorem**

Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.

For a positive integer n , the expansion is given by :

(a + x)^{n} = ^{n}C_{0}a^{n} + ^{n}C_{1}a^{n−1} x + ^{n}C_{2} a^{n−2} x^{2} + . . . + ^{n}C_{r} a^{n−r} x^{r} + . . . + ^{n}C_{n}x^{n}

where ^{n}C_{0}, ^{n}C_{1} , ^{n}C_{2} , . . . , ^{n}C_{n} are called Binomial co-efficients and there are (n + 1) terms in the expansion.

The value of ^{n}C_{r} is defined as :

**Similarly**,

(a−x)^{n} = ^{n}C_{0}a^{n} − ^{n}C_{1}a^{n−1} x + ^{n}C_{2} a^{n-2} x^{2} − . . . +( −1)^{r} ^{n}C_{r} a^{n−r} x^{r} + . . . + ( −1)^{n} ^{n}C_{n}x^{n}

**Illustration :** Expand

Solution:

^{7}C_{0}x^{7} + ^{7}C_{1}x^{6} + ^{7}C_{2}x^{5} + ^{7}C_{3}x^{4} + ^{7}C_{4}x^{3} + ^{7}C_{5}x^{2} + ^{7}C_{6} x + ^{7}C_{7}

= x^{7} + 7x^{5} + 21x^{3} + 35 x + 35/x + 21/x^{3} + 7/x^{5} + 1/x^{7}

**General Term in the Expansion:**

The general term in the expansion of (a + x)^{n} is the (r + 1)th term given as :

**t _{r+1} = ^{n}Cr a^{n-r} x^{r} .**

Similarly the general term in the expansion of (x + a)^{n} is given as :

**t _{r+1} = ^{n}C_{r} x^{n-r} a^{r} .**

The terms are considered from the beginning.

The (r + 1)th term from the end = (n − r + 1)th term from the beginning.

**Corollary: Coefficient of x ^{r} in expansion of (1 + x)^{n} is ^{n}C_{r}**

**Illustration :** Find the co-efficient of x^{24} in

**Solution:**

General term ((r + 1) th term) in

^{15}C_{r}(x^{2})^{15−r}(3a/x)^{r}

= ^{15}C_{r} x^{30-2r}(3^{r}a^{r}/x^{r})

= ^{15}C_{r} 3^{r}a^{r} x^{30-3r}

If this term contains x^{24} .

Then 30−3r = 24

=> 3r = 6 => r = 2

Therefore, the co-efficient of x^{24} = ^{15} C_{2} . 9a^{2}

Coefficients of Equidistant Terms from Beginning & end

The binomial coefficients in the expansion of (a + x)^{n} equidistant from the beginning and the end are equal.

i.e. ^{n}C_{r} = ^{n}C_{n−r}

**Note:**

∎

∎ If ^{n}C_{x} = ^{n}C_{y} , then either x = y or x + y = n

**Illustration 3:** If the co-efficient of (2r + 4)th term and (r − 2)th term in the expansion of (1 + x)^{18}are equal, find r

**Solution:** Since, co-efficient of (2r + 4)th term in (1 + x)^{18} = ^{18}C_{2r + 3 }

Co-efficient of (r − 2) th term = ^{18}C_{r − 3}

=> ^{18}C_{2r + 3} = ^{18}C_{r − 3}

=> 2r + 3 + r − 3 = 18 => 3r = 18

=> r = 6

**Illustration :** Find the last three digits of 17^{256}

Solution: We have 17^{2} = 289 =290 − 1

Now, 17^{256} = (17^{2})^{128} = (290 − 1)^{128}

= ^{128}C_{0}(290)^{128} − ^{128}C_{1}(290)^{127} + ……− ^{128}C_{125}(290)^{3} + ^{128}C_{126}(290)^{2} − ^{128}C_{127}(290) + 1

, where m is a positive integer.

= 1000 m + (128) (290) [(127) (145) − 1 ] + 1 = 1000 m + (128) (290) (18414) + 1

= 1000 [m + 683527] + 680 + 1 = 1000 [m + 683527] + 681

Thus, the last three digits of 17^{256} are 681