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.

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