HOW TO FIND COEFFICIENT OF X IN BINOMIAL EXPANSION

Example 1 :

Using binomial theorem, indicate which of the following two number is larger: (1.01)^1000000, 10000.

Solution :

  =  (1.01)^1000000 -  10000

  =  (1 + 0.01)^1000000 -  10000

= ^1000000C₀ + ^1000000C₁ + ^1000000C₂ + ..........  + ^1000000C1000000 (0.01)^1000000 - 10000

  =  (1 + 1000000 ⋅ 0.01 + other positive terms) - 10000

  =  (1 + 10000 + other positive terms) - 10000

  =  1 + other positive terms > 0

 0.01^1000000 >  10000

Example 2 :

Find the coefficient of x¹⁵ in  (x² + (1/x³))¹⁰

Solution :

General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x  =  x², n  =  10, a  =  1/x³

T_r+1  =  ⁿC_r x^(n-r) a^r

  =  ¹⁰C_r x^2(10-r) (1/x³)^r

  =  ¹⁰C_r x^20-2r (x^-3r)

  =  ¹⁰C_r x^20-5r ------(1)

Now let us find x¹⁵ th term

20 - 5r  =  15

20 - 15  =  5r

r  =  5/5  =  1

By applying the value of r in the (1)^st equation, we get

  =  ¹⁰C₁ x^20-5(1)

=  10

Hence the coefficient of  x¹⁵ is 10.

Example 3 :

Find the coefficient of x⁶ and the coefficient of x² in  (x² - (1/x³))⁶

Solution :

General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x  =  x², n  =  6, a  =  -1/x³

T_r+1  =  ⁿC_r x^(n-r) a^r

  =  ⁶C_r x^2(6-r) (-1/x³)^r

  =  ⁶C_r x^12-2r (-x^-3r)

  =  -⁶C_r x^12-5r ------(1)

x²  term

=  -⁶C₂ x^12-5(2)

=  -15x^12-10

=  -15 x²

Coefficient of x² term is -15.

Example 4 :

Find the coefficient of x⁴ in the expansion of (1 + x³)⁵⁰(x² + 1/x)⁵.

Solution :

  =  (1 + x³)⁵⁰ (x² + 1/x)⁵

  =  (1 + x³)⁵⁵/x⁵

  =  (1 + x³)⁵⁵/x⁵

General term  T_r+1  =  ⁿC_r x^(n-r) a^r

x  =  1, n  =  55, a  =  x³

T_r+1  =  ⁿC_r x^(n-r) a^r

  =  ⁵⁵C_r (1)^(55-r) (x³)^r

  =  ⁵⁵C_r x^3r/x⁵

  =  ⁵⁵C_r x^3r-5

3r - 5  =  4

3r  =  9

r  =  3

Coefficient of x⁴ is ⁵⁵C₃

  =  (55 ⋅ 54 ⋅ 53)/(3 ⋅ 2 ⋅ 1)

  =  26235

Hence the coefficient of x⁴ is 26235.

Example 5 :

Find the first 3 terms, in ascending powers of x, of the binomial expansion of

(2 - x/4)¹⁰

giving each term in the simplest form.

Solution :

= (2 - x/4)¹⁰

x = 2, a = -x/4, n = 10

Finding the first three terms,

= ¹⁰c₀ (2)^(10-0) (-x/4)⁰ + ¹⁰c₁ (2)^(10-1) (x/4)¹ + ¹⁰c₂ (2)^(10-2) (x/4)²

= ¹⁰c₀ (2)¹⁰ (-x/4)⁰ + ¹⁰c₁ (2)⁹ (x/4)¹ + ¹⁰c₂ (2)⁸ (x/4)²

= 1(2)¹⁰ (1) + 10(2)⁹ (x/4) + 45 (2)⁸ (x²/16)

= 1024 + 5120x/4 + 11520(x²/16)

= 1024 + 1280x + 720x²

Example 6 :

a)  Find the first 3 terms, in ascending powers of x, of the binomial expansion of

(2 - 3x)⁶

giving each term in the simplest form.

(b) Hence, or otherwise, find the first 3 terms, in ascending powers of x, of the expansion of

(1 + x/2)(2 - 3x)⁶

Solution :

a)

= (2 - 3x)⁶

x = 2, a = -3x, n = 6

Finding the first three terms,

= ⁶c₀ (2)^(6-0) (-3x)⁰ + ⁶c₁ (2)^(6-1) (-3x)¹ + ⁶c₂ (2)^(6-2) (-3x)²

= 1(2)⁶ (1) + 6(2)⁵ (-3x) + 15 (2)⁴ (9x²)

= 64 - 576x + 2160x²

b)  (1 + x/2)(2 - 3x)⁶

Writing the first three terms,

= (1 + x/2) (64 - 576x + 2160x²)

= 64 - 576x + 2160x² + 64x/2 - 576x (x/2) + 2160x² (x/2)

= 64 - 576x + 2160x² + 32x - 288x² + 1080x³ 

= 64 - 576x  + 32x + 2160x² - 288x² + 1080x³ 

= 64 - 544x + 1872x² + 1080x³ 

Constant, x term and x² terms are,

= 64 - 544x + 1872x²

Example 7 :

(a) Use the binomial theorem to find all the terms of the expansion of (2 + 3x)⁴ Give each term in its simplest form.

(b) Write down the expansion of (2 - 3x)⁴ in ascending powers of x, giving each term in its simplest form.

Solution :

a)

= (2 + 3x)⁴

x = 2, a = 3x, n = 4

Finding the first three terms,

= ⁴c₀ (2)^(4-0) (3x)⁰ + ⁴c₁ (2)^(4-1) (3x)¹ + ⁴c₂ (2)^(4-2) (3x)²

+ ⁴c₃ (2)^(4-3) (3x)³ + ⁴c₄ (2)^(4-4) (3x)⁴

= 1(2)⁴ (1) + 4(2)³ (3x) + 6(2)² (9x²) + 4(2)¹(27x³) + 1(2)⁰ (81x⁴)

= 1(2)⁴ (1) + 4(2)³ (3x) + 24 (9x²) + 8 (27x³) + 1 (81x⁴)

= 16 + 96x + 216x² + 216x³ + 81x⁴

b)

= (2 - 3x)⁴

x = 2, a = -3x, n = 4

Finding the first three terms,

= ⁴c₀ (2)^(4-0) (-3x)⁰ + ⁴c₁ (2)^(4-1) (-3x)¹ + ⁴c₂ (2)^(4-2) (-3x)²

+ ⁴c₃ (2)^(4-3) (-3x)³ + ⁴c₄ (2)^(4-4) (-3x)⁴

= 1(2)⁴ (1) + 4(2)³ (-3x) + 6(2)² (9x²) + 4(2)¹(-27x³) + 1(2)⁰ (81x⁴)

= 1(2)⁴ (1) - 4(2)³ (3x) + 24 (9x²) - 8 (27x³) + 1 (81x⁴)

= 16 - 96x + 216x² - 216x³ + 81x⁴

Example 8 :

Find the first 3 terms, in ascending powers of x, of the binomial expansion of (3+bx)⁵ where b is a non-zero constant. Give each term in its simplest form. Given that, in this expansion, the coefficient of x² is twice the coeffi cient of x, (b) find the value of b.

Solution :

= ⁵c₀ (3)^(5-0) (bx)⁰ + ⁵c₁ (3)^(5-1) (bx)¹ + ⁵c₂ (3)^(5-2) (bx)²

= 1 (3)⁵ (1) + 5(3)⁴ (bx)¹ + 10 (3)³ (b²x²)

= 243 + 405 bx + 270 b²x²

Coefficient of x² = 2 coefficient of x

270 b² = 2(405b)

270 b² = 810b

b = 810/270

b = 3

So, the value of b is 3.

Example 9 :

Given that ⁴⁰C₄ = 40!/4!b!

Write the value of b.

a)  In the binomial expansion of (1 + x)⁴⁰, the coefficient of x⁴ and x⁵ are p and q respectively.

b)  Find the value of q/p

Solution :

ⁿC_r = n!/n!(n-r)!

⁴⁰C₄= 40!/4!b!

= 40!/4!(40-4)!

b!= 36!

b = 36

a)  General term  T_r+1 = ⁿC_r x^(n-r) a^r

Finding coefficient of x⁴ :

x = 1, a = x, n = 40 and r = 4

T_r+1  =  ⁴⁰C₄ 1^(40-4) (x)⁴

= (40 ⋅ 39 ⋅ 38 ⋅ 37)/(4 ⋅ 3 ⋅ 2 ⋅ 1)1³⁶ x⁴

= (40 ⋅ 39 ⋅ 38 ⋅ 37)/(4 ⋅ 3 ⋅ 2 ⋅ 1)1³⁶ x⁴

= 91390 x⁴

Coefficient of x⁴ is 91390. So, the value of p is 91390

Finding coefficient of x⁵ :

x = 1, a = x, n = 40 and r = 5

T_r+1  =  ⁴⁰C₅ 1^(40-5) (x)⁵

= (40 ⋅ 39 ⋅ 38 ⋅ 37 ⋅ 36)/(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)1³⁶ x⁵

= (40 ⋅ 39 ⋅ 38 ⋅ 37)/(4 ⋅ 3 ⋅ 2 ⋅ 1)1³⁶ x⁵

= 658008 x⁵

Coefficient of x⁵ is 658008. So, the value of q is 658008.

b)

q/p = 658008/91390

= 7.2

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