If b is very small as compared to the value of a, so that the cube and other higher powers of b/a can be neglected in the identity 1/(a - b) + 1/(a - 2b) + 1/(a - 3b) + ..... + 1/(a - nb) = α n + β n² + γ n³, then the value of γ is :

Solution

Maths sir · 4 steps

Expand each fraction up to the b² term, then add: only the Σ r² part can make an n³.

1. Expand one term. Write the general term as 1/(a-rb) = 1/a(1-rb/a)⁻¹. Since b/a is small, expand binomially and drop cubes and higher powers.

   1/(a-rb) ≈ 1/a + b/a² r + b²/a³ r²

2. Add the terms. Summing from r = 1 to n gives n/a + b/a²·n(n+1)/2 + b²/a³·n(n+1)(2n+1)/6. The first two parts give at most n², so n³ comes only from the last part.
3. Pick out n cubed. n(n+1)(2n+1) = 2n³ + 3n² + n, so its n³ coefficient is 2. That gives γ = b²/a³·2/6.
4. So. γ = b²/3a³, which is option (c).

The idea to keep. When a small ratio is raised to the power r inside a sum, the power of r decides the power of n: Σ rᵏ grows like nᵏ⁺¹/(k+1). Match the highest power of n to the right term.

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