JEE Main 2023 · 30 January · morning · Single correct · +4 / −1
The coefficient of x³⁰¹ in (1 + x)⁵⁰⁰ + x(1 + x)⁴⁹⁹ + x²(1 + x)⁴⁹⁸ + ... + x⁵⁰⁰ is :
- (a) ⁵⁰⁰C₃₀₀
- (b) ⁵⁰¹C₂₀₀
- (c) ⁵⁰⁰C₃₀₁
- (d) ⁵⁰¹C₃₀₂
Answer and solution
Answer (b) ⁵⁰¹C₂₀₀
Solution
Maths sir · 4 steps
Each term is the one before it times x/(1+x), so sum it as a GP and almost everything collapses.
- Spot the GP. The terms are xᵏ(1+x)⁵⁰⁰⁻ᵏ for k = 0 to 500. That is 501 terms, with first term (1+x)⁵⁰⁰ and common ratio r = x/(1+x).
- Sum it. Here 1 - r = 1/(1+x), so dividing by 1 - r is the same as multiplying by (1+x). Also r⁵⁰¹ = x⁵⁰¹/(1+x)⁵⁰¹.
S = (1+x)⁵⁰⁰·(1 - r⁵⁰¹)/(1 - r) = (1+x)⁵⁰¹(1 - x⁵⁰¹/(1+x)⁵⁰¹) = (1+x)⁵⁰¹ - x⁵⁰¹
- Read the coefficient. The -x⁵⁰¹ term does not touch x³⁰¹. So the coefficient is the one in (1+x)⁵⁰¹, which is ⁵⁰¹C₃₀₁.
- So. ⁵⁰¹C₃₀₁ = ⁵⁰¹C₅₀₁₋₃₀₁ = ⁵⁰¹C₂₀₀, which is option (b).
The idea to keep. When the power of x goes up by one as the power of (1+x) goes down by one, you have a GP with ratio x/(1+x). It always sums to (1+x)ⁿ⁺¹ - xⁿ⁺¹.
Answer from the NTA final answer key. Solution by PrepWiser.