If the co-efficient of x⁷ and x⁸ in the expansion of ( 2 + x/3 )ⁿ are equal, then the value of n is equal to _____________.

Solution

Maths sir · 3 steps

Set the two coefficients equal and divide one by the other, so almost everything cancels.

1. General term. In (2 + x/3)ⁿ, the coefficient of xʳ is C(n, r) 2ⁿ⁻ʳ (1/3)ʳ.
2. Equate and divide. Set the coefficients of x⁷ and x⁸ equal: C(n, 7) 2ⁿ⁻⁷ 3⁻⁷ = C(n, 8) 2ⁿ⁻⁸ 3⁻⁸. Dividing, the leftover powers are 2ⁿ⁻⁷/2ⁿ⁻⁸ = 2 and 3⁻⁷/3⁻⁸ = 3.

   C(n, 8)/C(n, 7) = (n-7)/8 = 2 × 3 = 6

3. So. n - 7 = 48, which gives n = 55.

The common slip: 31. Dropping the powers of 2 leaves (n-7)/8 = 3, which gives 31, not 55.

The idea to keep. When two neighbouring coefficients are equal, take their ratio. Use C(n, r+1)/C(n, r) = (n-r)/(r+1) and keep every constant from both terms of the binomial.

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