Question

find the 11th term from the beginning and 11th terms end in the expansion of (2x-1/x^2)^25​plz ans its urgent

Answer 1

Answer:

²⁵C₁₀ * 2¹⁵ / x⁵

²⁵C₁₆ * 2⁹ / x²³

Step-by-step explanation:

(2x -  1/x²)²⁵

using (a + b)ⁿ

= ⁿC₀ aⁿb⁰  + ⁿC₁aⁿ⁻¹b¹  +  ⁿC₂aⁿ⁻²b²  + ....................+ ⁿCₙ₋₁a¹bⁿ⁻¹ + ⁿCₙa⁰bⁿ

a = 2x   ,  b =  -1/x²   n = 25

11th  term from beginning

= ²⁵C₁₀ (2x)²⁵⁻¹⁰ (-1/x²)¹⁰

= ²⁵C₁₀(2x)¹⁵ (-1/x²)¹⁰

= ²⁵C₁₀ * 2¹⁵ x¹⁵/x²⁰

=   ²⁵C₁₀ * 2¹⁵ / x⁵

11th term from end = 16th term from beginning

= ²⁵C₁₆ (2x)²⁵⁻¹⁶ (-1/x²)¹⁶

= ²⁵C₁₆(2x)⁹ (-1/x²)¹⁶

= ²⁵C₁₆ * 2⁹ x⁹/x³²

=   ²⁵C₁₆ * 2⁹ / x²³