FIND A SPECIFIC TERM IN A BINOMIAL EXPANSION

Problem 1:

Without simplifying, find:

The 6^th term of (2x + 5)¹⁵

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

To find 6^th term,

T₆ = T₅₊₁ of expansion (2x + 5)¹⁵

Putting r = 5, a = 2x, b = 5, n = 15

T₆ = ¹⁵C₅ (2x)^15-5 (5)⁵

T₆ = ¹⁵C₅ (2x)¹⁰ (5)⁵

Problem 2 :

The 4^th term of (x² + 5/x)⁹

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

To find 4^th term,

T₄ = T₃₊₁ of expansion (x² + 5/x)⁹

Putting r = 3, a = x², b = 5/x, n = 9

T₄ = ⁹C₃ (x²)^9-3 (5/x)³

T₄ = ⁹C₃ (x²)⁶ (5/x)³

Problem 3 :

The 10^th term of (x - 2/x)¹⁷

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

To find 10^th term,

T₁₀ = T₉₊₁ of expansion (x - 2/x)¹⁷

Putting r = 9, a = x, b = -2/x, n = 17

T₁₀ = ¹⁷C₉ (x)^17-9 (-2/x)⁹

T₁₀ = ¹⁷C₉ (x)⁸ (-2/x)⁹

Problem 4 :

The 9^th term of (2x² - 1/x)²¹

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

To find 9^th term,

T₉ = T₈₊₁ of expansion (2x² - 1/x)²¹

Putting r = 8, a = 2x², b = -1/x, n = 21

T₉ = ²¹C₈ (2x²)^21-8 (-1/x)⁸

T₉ = ²¹C₈ (2x²)¹³ (-1/x)⁸

Problem 5 :

Find the coefficient of:

x¹⁰ in the expansion of (3 + 2x²)¹⁰

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = 3, b = 2x², n = 10

T_r+1 = ¹⁰C_r (3)^10-r (2x²)^r

= ¹⁰C_r (3)^10-r 2^r x²^r

To find the coefficient of x¹⁰, take

x^2r = x¹⁰

2r = 10

r = 5

T₆ = ¹⁰C₅ (3)^10-5 2⁵ x²⁽⁵⁾

T₆ = ¹⁰C₅ 3⁵ 2⁵ x¹⁰

So, coefficient of x¹⁰ is

¹⁰C₅ 3⁵ 2⁵

Problem 6 :

x³ in the expansion of (2x² - 3/x)⁶

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = 2x², b = -3/x, n = 6

T_r+1 = ⁶C_r (2x²)^6-r (-3/x)^r

= ⁶C_r (2)^6-r (x²)^6-r (-3/x)^r

= ⁶C_r (2)^6-r (x)^12-2r (-3/x)^r

= ⁶C_r (2)^6-r (x)^12-2r (-3)^r (x)^-r

= ⁶C_r (2)^6-r (-3)^r (x)^12-3r

To find the coefficient of x³, take

x^12-3r = x³

12 - 3r = 3

-3r = 3 - 12

-3r = -9

r = 3

T₄ = ⁶C₃ (2)^6-3 (-3)³ x¹²^-9

T₄ = ⁶C₃ 2³ (-3)³ x³

So, coefficient of x³ is

⁶C₃ 2³ (-3)³

Problem 7 :

x¹² in the expansion of (2x² - 1/x)¹²

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = 2x², b = -1/x, n = 12

T_r+1 = ¹²C_r (2x²)^12-r (-1/x)^r

= ¹²C_r (2)^12-r (x²)^12-r (-1/x)^r

= ¹²C_r (2)^12-r (x)^24-2r (-1/x)^r

= ¹²C_r (2)^12-r (x)^24-2r (-1)^r (x)^-r

= ¹²C_r (2)^12-r (-1)^r (x)^24-3r

To find the coefficient of x¹², take

x^24-3r = x¹²

24 - 3r = 12

-3r = 12 - 24

-3r = -12

r = 4

T₅ = ¹²C₄ (2)^12-4 (-1)⁴ x²^4-12

T₅ = ¹²C₄ 2⁸ (-1)⁴ x¹²

So, coefficient of x¹² is

¹²C₄ 2⁸ (-1)⁴

Problem 8 :

Find the constant term in:

The expansion of (x + 2/x²)¹⁵

Solution:

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = x, b = 2/x², n = 15

T_r+1 = ¹⁵C_r (x)^15-r (2/x²)^r

= ¹⁵C_r (x)^15-r 2^r (x)^-2r

= ¹⁵C_r 2^r (x)^15-3r

Constant term:

15 - 3r = 0

-3r = -15

r = 5

T₆ = ¹⁵C₅ 2⁵ (x)^15-15

T₆ = ¹⁵C₅ 2⁵

So, the constant term is

¹⁵C₅ 2⁵

Problem 9 :

The expansion of (x - 3/x²)⁹

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = x, b = -3/x², n = 9

T_r+1 = ⁹C_r (x)^9-r (-3/x²)^r

= ⁹C_r (x)^9-r (-3)^r (x)^-2r

= ⁹C_r (-3)^r (x)^9-3r

Constant term:

9 - 3r = 0

-3r = -9

r = 3

T₄ = ⁹C₃ (-3)³ (x)^9-9

T₄ = ⁹C₃ (-3)³

So, the constant term is

⁹C₃ (-3)³