Question

If three consecutive coefficients in the expansion of (1 + x)ⁿ are in the ratio 6 : 33 : 110, find n.

Answer 1

Let the consecutive coefficients ⁿC_r, ⁿC_{r + 1} and ⁿC_{r + 2} be the coefficients of T_{r + 1}, T_{r + 2} and T_{r + 3} then ⁿC_r : ⁿC_{r + 1} : ⁿC_{r + 2} = 6 : 33 : 110

Now,

Subtracting (2) from (1), we get n = 12