Find the 11th term from the beginning and the 11th term from the end

1.	Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.
Answer» Given as (2x – 1/x²)²⁵ Now, the given expression contains 26 terms. Therefore, the 11^th term from the end is the (26 − 11 + 1) ^th term from the beginning. In other words, the 11^th term from the end is the 16^th term from the beginning. Now, T₁₆ = T₁₅₊₁ = ²⁵C₁₅ (2x)^25-15 (-1/x²)¹⁵ = ²⁵C₁₅ (2¹⁰) (x)¹⁰ (-1/x³⁰) = – ²⁵C₁₅ (2¹⁰ / x²⁰) Then we shall find the 11^th term from the beginning. T₁₁ = T₁₀₊₁ = ²⁵C₁₀ (2x)^25-10 (-1/x²)¹⁰ = ²⁵C₁₀ (2¹⁵) (x)¹⁵ (1/x²⁰) = ²⁵C₁₀ (2¹⁵ / x⁵)

Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.

Answer»

Given as

(2x – 1/x²)²⁵

Now, the given expression contains 26 terms.

Therefore, the 11^th term from the end is the (26 − 11 + 1) ^th term from the beginning.

In other words, the 11^th term from the end is the 16^th term from the beginning.
Now,

T₁₆ = T₁₅₊₁ = ²⁵C₁₅ (2x)^25-15 (-1/x²)¹⁵

= ²⁵C₁₅ (2¹⁰) (x)¹⁰ (-1/x³⁰)

= – ²⁵C₁₅ (2¹⁰ / x²⁰)

Then we shall find the 11^th term from the beginning.

T₁₁ = T₁₀₊₁ = ²⁵C₁₀ (2x)^25-10 (-1/x²)¹⁰

= ²⁵C₁₀ (2¹⁵) (x)¹⁵ (1/x²⁰)

= ²⁵C₁₀ (2¹⁵ / x⁵)