The coefficient of x⁵ in the expansion of (2 x³-1/(3 x²))⁵ is :

Solution

Maths sir · 4 steps

Write the general term, collect the powers of x, and set the exponent equal to 5 to find r.

1. General term. In (a+b)ⁿ the (r+1)th term is C(n, r)aⁿ⁻ʳbʳ. Here n = 5, a = 2x³ and b = -1/3x².

   Tᵣ₊₁ = C(5, r)(2x³)⁵⁻ʳ(-1/3x²)ʳ = C(5, r) 2⁵⁻ʳ(-1/3)ʳ x¹⁵⁻⁵ʳ

2. Find r. The power of x is 3(5-r) - 2r = 15 - 5r. Setting 15 - 5r = 5 gives r = 2, so the third term is the one we want.
3. Coefficient. Put r = 2: C(5, 2) × 2³ × (-1/3)² = 10 × 8 × 1/9. The square makes the sign positive.
4. So. The coefficient of x⁵ is 80/9, which is option (b).

The idea to keep. For a specific power of x, write Tᵣ₊₁ with every power of x gathered into one exponent, solve for r, then take the number in front. Watch the sign: an even power of a negative term is positive.

Answer from the NTA final answer key. Solution by PrepWiser.