Question

If 'n' is positive integer and three consecutive coefficient in the expansion of (1 + x)ⁿ are in the ratio 6 : 33 : 110, then n is equal to ______.

Options

• 9

• 6

• 12

• 16

Answer 1

If 'n' is positive integer and three consecutive coefficient in the expansion of (1 + x)ⁿ are in the ratio 6 : 33 : 110, then n is equal to 12.

Explanation:

Let the consecutive coefficient of

(1 + x)ⁿ are ⁿC_r–1, ⁿC_r, ⁿC_r+1

From the given condition,

ⁿC_r–1 : ⁿC_r : ⁿC_r+1 = 6 : 33 : 110

Now ⁿC_r–1 : ⁿC_r = 6 : 33

`\implies (n!)/((r - 1)!(n - r + 1)!) xx (r!(n - r)!)/(n!) = 6/33` 

`\implies r/(n - r + 1) = 2/11`

`\implies` 11r = 2n – 2r + 2

`\implies` 2n – 13r + 2 = 0  ...(i)

and ⁿC_r : ⁿC_{r + 1} = 33 : 110

`\implies` `(n!)/(r!(n - r)!) xx ((r + 1)!(n - r - 1)!)/(n!) = 33/110 = 3/10`

`\implies ((r + 1))/(n - r) = 3/10`

`implies` 3n – 13r – 10 = 0  ...(ii)

Solving (i) and (ii), we get n = 12