The coefficient of x¹⁰¹ in the expression (5 + x)⁵⁰⁰ + x(5 + x)⁴⁹⁹ + x²(5 + x)⁴⁹⁸ + ..... + x⁵⁰⁰, x > 0, is

Solution

Maths sir · 4 steps

This is a GP in disguise, so sum it first and the whole expression collapses to two terms.

1. Spot the GP. Each term is the previous one multiplied by x/(5+x). The first term is (5+x)⁵⁰⁰, and there are 501 terms, running from x⁰ up to x⁵⁰⁰.
2. Use the identity. The sum aⁿ + aⁿ⁻¹b + … + bⁿ equals (aⁿ⁺¹-bⁿ⁺¹)/(a-b). Here a = 5+x, b = x, n = 500, and a - b = 5.

   S = ((5+x)⁵⁰¹ - x⁵⁰¹)/((5+x) - x) = 1/5[(5+x)⁵⁰¹ - x⁵⁰¹]

3. Extract the coefficient. The x⁵⁰¹ term cannot give x¹⁰¹. In (5+x)⁵⁰¹, the x¹⁰¹ term is ⁵⁰¹C₁₀₁ 5⁴⁰⁰ x¹⁰¹, because 501 - 101 = 400.
4. So. Dividing by 5 gives the coefficient ⁵⁰¹C₁₀₁ 5³⁹⁹, which is option (a).

The idea to keep. When the powers of one bracket go down while the powers of x go up, you are looking at a GP. Use (aⁿ⁺¹-bⁿ⁺¹)/(a-b) and the long sum becomes one binomial.

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