The natural number m, for which the coefficient of x in the binomial expansion of
( xᵐ + 1/x² )²² is 1540, is .............

Solution

Maths sir · 4 steps

Use C(22, r) = 1540 to find r first, then the power of x fixes m.

1. General term. The (r+1)th term is C(22, r)(xᵐ)²²⁻ʳ(x⁻²)ʳ. Its coefficient is C(22, r) and its power of x is m(22-r) - 2r.

   Tᵣ₊₁ = C(22, r) xᵐ⁽²²⁻ʳ⁾⁻²ʳ

2. Match the coefficient. C(22, 3) = (22 × 21 × 20)/6 = 1540, and C(22, 19) is equal to it. So r = 3 or r = 19.
3. Match the power. For r = 3: 19m - 6 = 1 gives m = 7/19, which is not a natural number. For r = 19: 3m - 38 = 1 gives 3m = 39.
4. So. m = 13, so the answer is 13.

The idea to keep. When the coefficient is given, use it to find r first. Remember C(n, r) = C(n, n-r), so check both values of r against the power condition.

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