Question

What is eigth term of expansion when (2x²-1/2x²)

Answer 1

Answer:

$T₈ = C(n , 7)2ⁿ⁻¹⁴·x²⁽ⁿ⁻¹⁴⁾$

Step-by-step explanation:

$(2x² + \frac{1}{2x²})ⁿ$

The ᵢth term in the binomial expansion of (x + y)ⁿ is given by

$Tᵢ = C(n , i-1)xⁿ⁻ⁱ⁺¹ yⁱ⁻¹$

So, the eighth term is given by

$T₈ = C(n , 7)(2x²)ⁿ⁻⁷ (\frac{1}{2x²})⁷$

$T₈ = C(n , 7)2ⁿ⁻⁷·x²⁽ⁿ⁻⁷⁾·2⁻⁷· x⁻¹⁴$

$T₈ = C(n , 7)2ⁿ⁻¹⁴·x²⁽ⁿ⁻⁷⁾⁻¹⁴$

$T₈ = C(n , 7)2ⁿ⁻¹⁴·x²⁽ⁿ⁻¹⁴⁾$