HOW TO FIND EXPANSION USING BINOMIAL THEOREM

Binomial expansion for (x + a)ⁿ is,

nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + .........+ nc_nx^n-na⁰

Question 1 :

Expand (i) [2x² − (3/x)]³

Solution :

x  =  2x²,  a  =  (-3/x),  n  =  3

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  3c₀ (2x²)³(-3/x)⁰ + 3c₁ (2x²)²(-3/x)¹ + 3c₂ (2x²)¹(-3/x)² + 3c₃ (2x²)⁰(-3/x)³

  =  (8x⁶) + 3(4x⁴)(-3/x) + 3(2x²)(9/x²) + 1 (1)(-27/x³)

  =  8x⁶ - 36x³ + 54 - (27/x³)

Question 2 :

Expand (ii)  (2x² − 3√1 − x²)⁴  +  (2x² + 3√1 − x²)⁴

Solution :

Part 1 :

  =  (2x² − 3√1 − x²)⁴  

x  =  2x²,  a  =  (-3√1 − x²),  n  =  4

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  4c₀ (2x²)⁴(-3√1 −x²)⁰

+ 4c₁ (2x²)³(-3√1−x²)¹ + 4c₂ (2x²)²(-3√1−x²)²

+ 4c₃ (2x²)¹(-3√1 −x²)³+ 4c₄ (2x²)⁰(-3√1 −x²)⁴

=  16x⁸

+ 4 (8x⁶)(-3√1−x²)¹ + 6 (4x⁴) (9√1−x²)²

+ 4c₃ (2x²)¹(-27√1 −x²)³+ 4c₄ (2x²)⁰(-3√1 −x²)⁴

  =  16x⁸

- 96x⁶ √1−x² + 216x⁴ √1−x² - 216x²(1-x²) √1−x² 

  + 81 (1 −x²)²  --------(1)

Part 2 :

  =  (2x² + 3√1 − x²)⁴  

x  =  2x²,  a  =  (3√1 − x²),  n  =  4

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  4c₀ (2x²)⁴(3√1 −x²)⁰

+ 4c₁ (2x²)³(3√1−x²)¹ + 4c₂ (2x²)²(3√1−x²)²

+ 4c₃ (2x²)¹(3√1 −x²)³+ 4c₄ (2x²)⁰(3√1 −x²)⁴

  =  16x⁸

+ 96x⁶ √1−x² + 216x⁴ √1−x² + 216x²(1-x²) √1−x² 

  + 81 (1 −x²)²  --------(2)

(1) + (2)  ==>  2 [16x⁸ + 216x⁴ √1−x² + 81 (1 −x²)²]

Hence the answer is 2 [16x⁸ + 216x⁴ √1−x² + 81 (1 −x²)²].

Question 3 :

Compute (i)  102⁴

Solution :

102⁴  =  (100 + 2)⁴

x  =  100, a  =  2, n  =  4

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  4c₀ (100)⁴(2)⁰ + 4c₁(100)³(2)¹ + 4c₂(100)²(2)² + 4c₃(100)¹(2)³ + 4c₄(100)⁰(2)⁴

  =  1 (100000000)(1) + 4(1000000)(2) + 6(10000)(4) + 4(100)¹(8) + 1(1)(16)

  =  100000000 + 8000000 + 240000 + 3200 + 16

  =  108243216

Hence the value of  102⁴ is 108243216.

Question 4 :

Compute (ii)  99⁴

Solution :

99⁴  =  (100 - 1)⁴

x  =  100, a  =  -1, n  =  4

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  4c₀ (100)⁴(-1)⁰ + 4c₁(100)³(-1)¹ + 4c₂(100)²(-1)² + 4c₃(100)¹(-1)³ + 4c₄(100)⁰(-1)⁴

  =  1 (100000000)(1) + 4(1000000)(-1) + 6(10000)(1) + 4(100)¹(-1) + 1(1)(1)

  =  100000000 - 4000000 + 60000 - 400 + 1

  =  96059601

Hence the value of  99⁴ is 96059601.

Question 5 :

Compute (iii)  9⁷

Solution :

9⁷  =  (10 - 1)⁷

x  =  10, a  =  -1, n  =  7

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

=  7c₀ (10)⁷(-1)⁰ + 7c₁(10)⁶(-1)¹ + 7c₂(10)⁵(-1)² + 7c₃(10)⁴(-1)³ + 7c₄(10)³(-1)⁴+ 7c₅(10)²(-1)⁵+ 7c₆(10)⁰(-1)⁶  + 7c₇(10)¹(-1)⁷

  =  1(10000000) - 7(1000000) + 21(100000) - 35(10000) + 35(1000) - 21(100) + 7 - 1

  =  10000000-7000000+2100000-350000+35000- 2100+70-1

=  4782969 

Hence the value of  9⁷ is 4782969.

Question 6 :

Find the coefficient of x¹¹ in the expansion of

(x³ - 2/x²)¹²

Solution :

(x³ - 2/x²)¹²

General term

T_{r + 1} = ⁿC_r x^n-r a^r

n = 12, x = x³  a = - 2/x²

= ¹²C_r (x³)^12-r (- 2/x²)^r

Coefficient of x¹¹

= ¹²C_r (x^36-3r)(- 2^r/x^2r)

= -2^r (¹²C_r) (x^36-3r)(x^2r)

= -2^r (¹²C_r) (x^36-3r-2r)

= -2^r (¹²C_r) (x^36-5r) --------(1)

36 - 5r = 11

36 - 11 = 5r

25 = 5r

r = 5

To find coefficient of  x¹¹, we apply r = 5

= -2⁵ (¹²C₅) (x^36-5(5))

= -32 (12 x 11 x 10 x 9 x 8/5 x 4 x 3 x 2 x 1)/ (x¹¹)

= -25344x¹¹

So, the required coefficient of x¹¹ is -25344.

Question 7 :

Determine whether the expansion of 

(x² - 2/x)¹⁸

will contain a term containing x¹⁰ ?

Solution :

(x² - 2/x)¹⁸

General term

T_{r + 1} = ⁿC_r x^n-r a^r

n = 18, x = x²  a = - 2/x

= ¹⁸C_r (x²)^18-r (-2/x)^r

Coefficient of x¹⁰

= ¹⁸C_r (x²)^18-r (-2^r/x^r)

= ¹⁸C_r (x²)^18-r x^r(-2^r)

= ¹⁸C_r (x^{36 - 2r + r})(-2^r)

= ¹⁸C_r (x^{36 - 3r})(-2^r)

36 - 3r = 10

3r = 36 - 10

3r = 26

r = 26/3

Since r is the fraction, by dividing it we get decimal. The position cannot be decimal. Then the expansion cannot contain x¹⁰.

Question 8 :

The total number of terms in the expansion of (x + a)⁵¹– (x – a)⁵¹ after simplification is

(a) 102      (b) 25      (c) 26      (d) None of these

Solution :

(x + a)⁵¹– (x – a)⁵¹

As a expansion of first binomial, we will get 52 terms in total and expanding the second binomial.

Since we have negative after that and getting 52 terms as the expansion of the second binomial. Half of the terms can be eliminated. So, the total number of terms after the simplification is 26.

Question 9 :

The coefficient of x² of the expansion (d - x)³ is -9. Find the value of d.

Solution :

(d - x)³ 

Here n = 3, x = d and a = -x

Finding coefficient of x² separately.

T_{r + 1} = ⁿC_r x^n-r a^r

T_{r + 1} = ³C_r d^3-r (-x)^r

= ³C_r d^3-r (-x)^r

coefficient of x² term, we put r = 2

³C₂ d^3-2 (-x)²

The given coefficient of x² term is -9, then

3d = -9

d = -3

Question 10 :

Find the expansion of (1 + 3x)(1 - 2x)⁴

Solution :

(1 - 2x)⁴

Finding the expansion, we get

= nc₀xⁿa⁰ + nc₁x^n-1a¹ + nc₂x^n-2a² + ........+ nc_nx^n-na⁰

Here n = 4, x = 1 and a = -2x

= 4c₀(1)⁴(-2x)⁰ + 4c₁(1)^4-1(-2x)¹ + 4c₂(1)^4-2(-2x)² + 4c₃(1)^4-3(-2x)³ + 4c₄(1)^4-4(-2x)⁴ 

= 1 + 4(-2x) + 6(4x²) + 4(-8x³) + 1(16x⁴)

= 1 - 8x + 24x² - 32x³ + 16x⁴

(1 + 3x)(1 - 2x)⁴

= (1 + 3x) (1 - 8x + 24x² - 32x³ + 16x⁴)

= 1 - 8x + 24x² - 32x³ + 16x⁴ + 3x - 24x² + 72x³ - 96x⁴ + 48x⁵

= 48x⁵ + 16x⁴ - 96x⁴ - 32x³ + 72x³ + 24x² - 24x² - 8x + 3x + 1

= 48x⁵ - 80x⁴  + 40x³ - 5x + 1

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