General Term & Middle Term of Binomial Expansion | Term Independent of X & Greatest Term in Binomial Expansion | Solved Examples

📚 General Term & Middle Term of Binomial Expansion | Term Independent of X & Greatest Term – Class 11 Math

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General Term of Binomial Expansion

The general term of the binomial expansion signifies a term that produces all the terms of the binomial expansion by simply replacing the value of one component of the term.

Consider the binomial expansion, (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1 b + ⁿC₂ a^n-2 b² + … + ⁿC_n-1 a b^n-1 + ⁿC_n bⁿ

Here,

First term = ⁿC₀ aⁿ

Second term = ⁿC₁ a^n-1 b

Third term = ⁿC₂ a^n-2 b²

Similarly, we can write the (r + 1)th term as:

ⁿC_r a^n-r b^r

This is the general term of the given expansion.

Thus, (r + 1)Th term, i.e. T_r+1 = ⁿC_r a^n-r b^r is called the middle term of the expansion (a + b)ⁿ.

For example in the binomial expansion of (x+y)ⁿ the general term is,

T_r+1 = ⁿC_r x^n-ry^r

Where, 0 ≤ r ≤n

Substituting the values of r in the above term we get all the terms of the expansion. We observe that r takes n+1 values which is also true as there are n+1 terms in the expansion of (x+y)ⁿ

Example.1 : Find the 6^th term in the expansion of (3x + 4)⁸

Solution:

The general term in the expansion of (3x + 4)⁸ is,

T_r+1 = ⁸C_r(3x)^8-r(4)^r

For 6^th term r = 5

T₆ = T₅₊₁ = ⁸C₅(3x)^8-5(4)⁵

Simplifying the above term we get our answer,

T₆ = ⁸C₅(3x)^8-5(4)⁵

= (8.7.6/3.2.1)(27)(1024)x³

= 1593648x³

Example 2: Find the 3^th term in the expansion of (2x + 1)⁴

Solution:

The general term in the expansion of (2x + 1)⁴ is,

T_r+1 = ⁴C_r(2x)^4-r(1)^r

For 3^th term r = 2

T₃ = T₂₊₁ = ⁴C₂(2x)^4-2(1)²

Simplifying the above term we get our answer,

T₃ = ⁴C₂(2x)^4-2(1)²

= (6)(4)x²

= 24x²

Example 3: Find the 4th term from the end in the expansion of ((x³/2) – (2/x²))⁸.

Solution:

Since r^th term from the end in the expansion of (a + b)ⁿ is (n – r + 2)^th term from the beginning.

Therefore, 4^th term from the end is 8 – 4 + 2,

i.e., 6th term from the beginning, which is given by

T₆ = ⁸C₅(x³/2)³(-2/x²)⁵

= ⁸C₃(x⁹/8)(-32/x¹⁰)

= -224/x

Example.4 : Find the fourth term from the end in the expansion of (2x – 1/x²)¹⁰

Solution :

Required term =T_{10 – 4 + 2} = T₈ = 10C₇ (2x)³ (−1/x²)⁷ = −960x^-11

Example 5 : Find the coefficient of x⁶y³ in the expansion (x + 2y)⁹.

Solution:

Let x⁶y³ be the (r + 1)th term of the expansion (x + 2y)⁹.

So,

T_r+1 = ⁹C_r x^9-r (2y)^r

x⁶y³ = ⁹C_r x^9-r 2^r y^r

By comparing the indices of x and y, we get r = 3.

Coefficient of x⁶y³ = ⁹C₃ (2)³

= 84 × 8

= 672

Therefore, the coefficient of x⁶y³ in the expansion (x + 2y)⁹ is 672.

Example 6: The second, third and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720 and 1080, respectively. Find x, a and n.

Solution:

Given,

Second term = T₂ = 240

Third term = T₃ = 720

Fourth term = T₄ = 1080

Now,

T₂ = T₁₊₁ = ⁿC₁ x^n-1 (a)

ⁿC₁ x^n-1 a = 240….(i)

Similarly,

ⁿC₂ x^n-2 a² = 720….(ii)

ⁿC₃ x^n-3 a³ = 1080….(iii)

Dividing (ii) by (i),

[ⁿC₂ x^n-2 a²]/ [ⁿC₁ x^n-1 a] = 720/240

[(n – 1)!/(n – 2)!].(a/x) = 6

(n – 1) (a/x) = 6

a/x = 6/(n – 1)….(iv)

Similarly, by dividing (iii) by (ii),

a/x = 9/[2(n – 2)]….(v)

From (iv) and (v),

6/(n – 1) = 9/[2(n – 2)]

12(n – 2) = 9(n – 1)

12n – 24 = 9n – 9

12n – 9n = 24 – 9

3n = 15

n = 5

Subsituting n = 5 in (i),

⁵C₁ x⁴ a = 240

ax⁴ = 240/5

ax⁴ = 48….(vi)

Substituting n = 5 in (iv),

a/x = 6/(5 – 1)

a/x = 6/4 = 3/2

a = (3x/2)

Putting this oin equ (vi), we get;

(3x/2) x⁴ = 48

x⁵ = 32

x⁵ = 25

⇒ x = 2

Substituting x = 2 in a = (3x/2)

a = 3(2)/2 = 3

Therefore, x = 2, a = 3 and n = 5.

Question 7: The coefficients of three consecutive terms of (1 + x)ⁿ⁺⁵ are in the ratio 5:10:14. Find n.

Solution:

Let T_r-1, T_r, T_r+1 are three consecutive terms of (1 + x)ⁿ⁺⁵

⇒  T_r-1 = (n+5) C_r-2 . x^r-2

⇒  T_r = (n+5) C_r-1 . x^r-1

⇒  T_r+1 = (n+5) C_r . x^r

Given

(n+5) C_r-2 : (n+5) C_r-1 : (n+5) C_r = 5 : 10 : 14

Therefore, [(n+5) C_r-2]/5= [(n+5) C_r-1]/10 = (n+5) C_r/14

Comparing first two results we have n – 3r = -9 . . . . . . (1)

Comparing last two results we have 5n – 12r = -30 . . . . . .  (2)

From equations (1) and (2), n = 6

Middle Term of Binomial Expansion

We know that the total number of terms in the expansion of (x + y)ⁿ is n + 1. And so the middle term in the binomial expansion of the (x+y)ⁿ depends on the value of n. The value of n can be either even or odd which decides the number/numbers and values of the middle term.

• If n is even then we have odd (n+1) terms in the expansion of (x+y)ⁿ. Thus, there is one middle term, in this case, and the middle term is (n/2 + 1)^th term.
• If n is odd then we have even (n+1) terms in the expansion of (x+y)ⁿ. Thus, there are two middle terms, in this case, and the middle terms are the (n/2)^th term and (n/2 + 1)^th term.

Example 1: Find the middle term (terms) in the expansion of ((p/x) + (x/p))⁹

Solution:

Since the power of binomial is odd. Therefore, we have two middle terms which are 5th and 6th terms. These are given by

T₅  = ⁹C₄(p/x)⁵(x/p)⁴

= ⁹C₄(p/x)

= 126(p/x)

T₆  = ⁹C₅(p/x)⁴(x/p)⁵

= ⁹C₅(x/p)

= 126(x/p)

Example.2 : Find the middle term of (1 −3x + 3x² – x³)²ⁿ

Solution :

(1 − 3x + 3x² – x³)²ⁿ = [(1 − x)³]²ⁿ = (1 − x)⁶ⁿ

Middle Term = [(6n/2) + 1] term = 6nC_3n (−x)³ⁿ

Example3 : The sum of the real values of x for which the middle term in the binomial expansion of (x³/3  + 3/x)⁸ equals 5670 is?

Solution:

T₅ = ⁸C₄ × (x¹²/81) × (81/x⁴) = 5670

⇒ 70 x⁸ = 5670

⇒ x = ± √3.

Identifying a Particular Term in Expansion

Any particular term can easily be identified in the expansion of (x+y)ⁿ. We follow two steps to get a particular term in the expansion of (x+y)ⁿ.

For example, if we have to find the term p^th term in the expansion of (x+y)ⁿ.

Step 1: Find the general term in the expansion of (x+y)ⁿ. The general term is,

T_r+1 = ⁿC_r x^n-ry^r

Step 2: Then substitute the value of ‘r’ for the value of the term which we have to find. In this case, r = p and then simplify to get the p^th term.

Determining a Particular Term:

• In the expansion of (ax^p + b/x^q)ⁿ the coefficient of x^m is the coefficient of T_r+1 where r = [(np−m)/(p+q)]
• In the expansion of (x + a)ⁿ, T_r+1/T_r = (n – r + 1)/r . a/x

Term Independent of X in Expansion

The term independent of ‘x’ can easily be identified in the expansion of (x+y)ⁿ. We follow two steps to get a particular term in the expansion of (x+y)ⁿ.

For example, if we have to find the term independent of ‘x’ term in the expansion of (x+y)ⁿ.

Step 1: Find the general term in the expansion of (x+y)ⁿ. the general term is,

T_r+1 = ⁿC_r x^n-ry^r

Step 2: Then substitute the value of ‘r’  as r = n to get rid of the x term.

The term Independent of in the expansion of [ax^p + (b/x^q)]ⁿ is

T_r+1 = ⁿC_r a^n-r b^r, where r = (np/p+q) (integer)

Example.1: Find the term independent of x in the expansion of (2x + 1)⁸

Solution:

The general term in the expansion of (2x + 1)⁸ is,

T_r+1 = ⁸C_r(2x)^8-r(1)^r

For the term independent of x

8-r = 0

r = 8

T₉ = T₈₊₁ = ⁸C₈(2x)^8-8(1)⁸

Simplifying the above term we get our answer,

T₉ = ⁸C₈(2x)^8-8(1)⁸

T₉ = (1)(1)(1) = 1

Example 2: Find the term independent of x in the expansion of (3x + 2)⁸

Solution:

The general term in the expansion of (2x + 1)⁸ is,

T_r+1 = ⁸C_r(3x)^8-r(2)^r

For the term independent of x

8-r = 0

r = 8

T₉ = T₈₊₁ = ⁸C₈(3x)^8-8(2)⁸

Simplifying the above term we get our answer,

T₉ = ⁸C₈(3x)^8-8(2)⁸

= (1)(1)(256)

= 256

Example.3 : Find the independent term of x in (x+1/x)⁶

Solution :

r = [6(1)/1+1] = 3

The independent term is 6C₃ = 20

Numerically Greatest Term in the Expansion of (1+x)ⁿ:

• If [(n+1)|x|]/[|x|+1] = P, is a positive integer, then the P^th term and (P+1)^th terms are numerically the greatest terms in the expansion of (1+x)ⁿ
• If[(n+1)|x|]/[|x|+1] = P + F, where P is a positive integer, and 0 < F < 1, then (P+1)^th term is numerically the greatest term in the expansion of (1+x)ⁿ.

While using this formula we have to make sure that the expansion is in (1+x)ⁿ form only, and |x| gives only the numerical value.

Example.1 : Find the numerically greatest term in (1-3x)¹⁰ when x = (1/2)

Sol:ution :

[(n + 1)|α|] / [|α| + 1] = (11 × 3/2)/(3/2+1) = 33/5 = 6.6

Therefore, T₇ is the numerically greatest term.

T_{6 + 1} = 10C₆ . (−3x)⁶ = 10C₆ . (3/2)⁶

Ratio of Consecutive Terms/Coefficients:

The coefficients of x^r and x^{r + 1} are ⁿC_{r – 1} and ⁿC_r, respectively.

(ⁿC_r / ⁿC_{r – 1}) = (n – r + 1) / r

Example.1 : If the coefficients of three consecutive terms in the expansion of (1+x)ⁿ are in the ratio 1:7:42, then find the value of n.

Solution :

Let (r – 1)^th, (r)^th and (r + 1)^th be the three consecutive terms.

Then, the given ratio is 1:7:42

Now (ⁿC_r-2 / ⁿC_{r – 1}) = (1/7)

(ⁿC_r-2 / ⁿC_{r – 1}) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)

And,

(ⁿC_r-1 / ⁿC_r) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)

From (1) & (2), n = 55

Practice Problems

1. The coefficients of three consecutive terms in the expansion of (1 + a)ⁿ are in the ratio1: 7 : 42. Find n.
2. What is the middle term in the expansion of [3x – (x³/6)]⁷?
3. If the coefficients of a^r-1, a^r and a^r+1 in the expansion of (1 + a)ⁿ are in arithmetic progression, prove that n² – n(4r + 1) + 4r² – 2 = 0.

🔢 Understanding Binomial Expansion

The Binomial Theorem provides a systematic way to expand expressions of the form (a + b)ⁿ. In this post, we will cover:

✅ General Term in Binomial Expansion
✅ Middle Term in Binomial Expansion
✅ Term Independent of x
✅ Greatest Term in Binomial Expansion

These concepts are crucial for CBSE Class 11, JEE Mains, and JEE Advanced, helping students solve algebraic expressions quickly and efficiently.

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