If the second, third and fourth terms in the expansion of (x + y)n are 135, 30 and 103, respectively, then 6 (n3 + x2 + y ) is equal to .
Answer
806
Solution
T2 = nC1xn-1y = 135, T3 = nC2xn-2y2 = 30, T4 = nC3xn-3y3 = 103
T3T2 = n - 12 yx = 29 ...(i)
T4T3 = n - 23 yx = 19 ...(ii)
(i) ÷ (ii): 3(n - 1)2(n - 2) = 2 n = 5; then (i) gives yx = 19, i.e. x = 9y
T2: 5x4 x9 = 135 x5 = 243 x = 3, y = 13
6 (n3 + x2 + y ) = 6 (125 + 9 + 13 ) = 806
Answer: 806