Binomial theorem or expansion describes the algebraic expansion of powers of a binomial. According to this theorem, it is possible to expand the polynomial “**(a + b) ^{n}**

^{“}into a sum involving terms of the form “

**ax**

^{z}y^{c}^{“}, the exponents z and c are non-negative integers where z + c = n, and the coefficient of each term is a positive integer depending on the values of n and b.

Example: If n = 4

(a + b)^{4 }= a^{4 }+ 4a^{3}y + 6a^{2}b^{2 }+ 4ab^{3 }+ b

**General Term of Binomial Expansion**

The General Term of Binomial Expansion of (x + y)^{n} is as follows

- T
_{r+1 }is the General Term in the binomial expansion - The General term expansion is used to find the terms mentioned in the above formula.
- To find the terms in the binomial expansion we need to expand the given expansion.
- Suppose (a + b)
^{n }is the equation then the series of its binomial expansion will be as follows:

- The first term of the series is T
_{1}=^{n}C_{0}.a^{n} - The second term of the series is T
_{2}=^{n}C_{1}.a^{n-1}.b - The third term of the series is T
_{3}=^{n}C_{2}.a^{n-2}.b^{2} - The n
^{th}term of the series is T_{n}=^{n}C_{n}.b^{n}

### Sample Problems on General Term

**Example 1: Find (r+1) ^{th }term for the given binomial expansion (x + 2y)^{5}**

**Solution: **

Given expansion is (x + 2y)

^{5}a = x, b= 2y, n = 5

The formula for (r+1)

^{th}is^{n}C_{r}.a^{n – r}.b^{r}(r+1)

^{th }term =^{5}C_{r}.x^{5 -r}.2y^{r}

**Example 2: Find (r+1) ^{th} term for the given binomial expansion (a + 2b)^{7}**

**Solution:**

Given expansion is (a + 2b)

^{7}a = a, b = 2b, n = 7

The formula for (r+1)th is

^{n}C_{r}.a^{n – r}.b^{r}(r+1)

^{th}term =^{7}C_{r}.a^{7 -r}.b^{r}

**Example 3: Find (r + 1) ^{th} term for the given binomial expansion (6p + 2q)^{12}**

**Solution: **

Given expansion is (6p + 2q)

a = 6p, b = 2q, n = 12

The formula for (r+1)

^{th}is^{n}C_{r}.a^{n – r}.b^{r}(r + 1)

^{th }term =^{12}C_{r}.6p^{12 -r}.2q^{r}

**Middle Term of the Binomial Expansion**

If **(x + y) ^{n }= ^{n}C_{r}.x^{n – r.}y^{r}**

^{ }, it has (n + 1) terms and the middle term will depend upon the value of n.

We have two cases for the Middle Term of a Binomial Expansion:

**If n is Even **

If n is the even number then we make it into an odd number and consider (n + 1) as odd and (n/2 + 1) is the middle term. In simple, if n is even then we consider it as odd.

Suppose n is the even so, (n + 1) is odd. To find out the middle term :

Consider the general term of binomial expansion i.e.

- Now we replace “r ” with “n/2” in the above equation to find the middle term
- T
_{r+1}= T_{n/2}+ 1 - T
_{n/2 + 1}=^{n}C_{n/2}.x^{n – n/2}.y^{n/2}

### Sample Problems on Middle Terms

**Example 1: Find the middle of the following binomial expansion (x + a) ^{8}**

**Solution:**

Given expansion is (x + a)

^{8}n = 8, we consider the expansion has (n + 1) terms so the above expansion has (8 + 1) i.e 9 terms

we have T

_{1}, T_{2}, T_{3}, T_{4}, T_{5}, T_{6}, T_{7}, T_{8}, T_{9}.T

_{r+1}= T_{n/2 + 1}=^{n}C_{n/2}.x^{n – n/2}.Y^{n/2}T

_{8/2 + 1}=^{8}C_{8/2}.x^{8-8/2}.a^{8/2}T

_{5}=^{8}C_{4}.x^{4}.a^{4}is the required middle term of the given binomial expansion.

**Example 2: Find the middle of the following binomial expansion (x + 3y) ^{6}**

**Solution:**

Given expansion is (x + 3y)

^{6}n = 6, we consider the expansion has (n + 1) terms so the above expansion has (6 + 1) i.e 7 terms

we have T

_{1}, T_{2}, T_{3}, T_{4}, T_{5}, T_{6}, T_{7}.T

_{r+1}= T_{n/2 + 1}=^{n}C_{n/2}.x^{n – n/2}.Y^{n/2}T

_{6/2 + 1}=^{6}C_{6/2}.x^{6-6/2}.3y^{6/2}T

_{4}=^{6}C_{3}.x^{3}.3y^{3}is the required middle term of the given binomial expansion.

**Example 3: **Find the middle of the following binomial expansion (2x + 5y)^{4}

**Solution: **

Given expansion is (2x – 5y)

^{4}n = 4, we consider the expansion has (n + 1) terms so the above expansion has (4 + 1) i.e 5 terms

we have T

_{1}, T_{2}, T_{3}, T_{4}, T_{5}.T

_{r+1}= T_{n/2 + 1 }=^{n}C_{n/2}.x^{n – n/2}.Y^{n/2}T

_{4/2 + 1}=^{4}C_{4/2}.2x^{4-4/2}.5y^{4/2}T

_{3}=^{4}C_{2}.x^{2}.5y^{2}is the required middle term of the given binomial expansion.

### If n is Odd

If n is the odd number then we make it into an even number and consider (n + 1) as even and (n + 1/2), (n + 3/2) is the middle terms. In simple, if n is odd then we consider it as even.

We have two middle terms if n is odd. To find the middle term:

Consider the general term of binomial expansion i.e

- In this case, we replace “r” with the two different values
- One term is (n + 1/2) compare with (r + 1) terms we get

r + 1 = n + 1/2

r = n + 1/2 -1

r = n -1/2

- Second middle term , compare (r + 1) with (n + 3/2) we get

r +1 = n +3/2

r = n + 3/2 – 1

r = n + 1/2

The two middle terms when n is odd are **(n – 1/2)** and **(n + 1/2)**.

### Sample Problems on Middle Terms

**Example 1: Find the middle terms of the following binomial expansion (x + a) ^{9}**

**Solution: **

Given expansion is (x + a)

^{9}a = x, b = a, and n = 9

Middle terms will be (n – 1)/2 and (n + 1)/2

T

_{r + 1 }=T_{n – 1/2}and T_{n + 1/2}First middle term:

T

_{r + 1}= T_{n – 1/2}=^{9}C_{(n – 1)/2}.x^{9 – (n – 1)/2}.a^{(n – 1)/2}T

_{(9 – 1/2)}=^{9}C_{(9 – 1)/2}.x^{9 – (n – 1)/2}.a^{(n – 1)/2}T

_{4}=^{9}C_{4}.x^{5}.a^{4}Second middle term:

T

_{r + 1}= T_{n + 1/2}=^{9}C_{(n + 1)/2}.x^{9 – (n + 1)/2}.a^{(n + 1)/2}T

_{(9 + 1)/2}=^{9}C_{(9 + 1)/2}.x^{9 – (9 + 1)/2}.a^{(9 + 1)/2}T

_{5 }=^{9}C_{5}.x^{4}.a^{5}The middle terms of expansion are T

_{4}, and T_{5.}

**Example 2: Find the middle terms of the following binomial expansion (4a + 9b) ^{7}**

**Solution: **

Given expansion is (4a + 9b)

^{7}a = 4a,b = 9b, and n = 7

middle terms will be (n -1/2) and (n + 1/2)

T

_{r + 1}=T_{n – 1/2}and T_{n + 1/2}First middle term:

T

_{r + 1}= T_{n – 1/2}=^{7}C_{(n – 1)/2}.4a^{7 – (n – 1)/2}.9b^{(n – 1)/2}T

_{(7 – 1/2) }=^{7}C_{(7 – 1)/2}.4a^{7 – (7 – 1)/2}.9b^{(7 – 1)/2}T

_{3}=^{7}C_{3}.4a^{4}.9b^{3}Second middle term:

T

_{r + 1}= T_{(n + 1)/2 }=^{7}C_{(n + 1)/2}.4a^{7 – (n + 1)/2}.9b^{(n + 1)/2}T

_{(7 + 1)/2}=^{7}C_{(7 + 1)/2}.4a^{7 – (7 + 1)/2}.9b^{(7 + 1)/2}T

_{4}=^{7}C_{4}.4a^{3}.9b^{4}The middle terms of expansion are T

_{3}, and T_{4}.^{}

**Example 3: Find the middle terms of the following binomial expansion (2x + 8y) ^{5}**

**Solution:**

Given expansion is (2x + 8y)

^{5}a = 2x,b = 8y, and n = 5

Middle terms will be (n – 1)/2 and (n + 1)/2

T

_{r + 1}= T_{(n – 1)/2}and T_{(n + 1)/2}First middle term:

T

_{r + 1}= T_{(n – 1)/2}=^{5}C_{(n – 1)/2}.2x^{5 – (n – 1)/2}.8y^{(n – 1)/2}T

_{(5 – 1)/2}=^{5}C_{(5 – 1)/2}.2x^{5 – (5 – 1)/2}.8y^{(5 – 1)/2}T

_{2}=^{5}C_{2}.2x^{3}.8y^{2}Second middle term:

T

_{r + 1}= T_{(n + 1)/2}=^{5}C_{(n + 1)/2}.2x^{5 – (n + 1)/2}.8y^{(n + 1)/2}T

_{(5 + 1)/2}=^{5}C_{(5 + 1)/2}.2x^{5 – (5 + 1)/2}.8y^{(5 + 1)/2}T

_{3}=^{5}C_{3}.2x^{2}.8y^{3}The middle terms of expansion are T

_{2}, and T_{3}.