Question

If n is positive integer and three consecutive

coefficients in the expansion of $(1+x{)}^n$ are in the

ratio 6 : 33 : 110, then n =

• A. 4
• B. 6
• C. 12
• D. 16

Answer 1

The correct option is C

12

Let the consecutive coefficient of $(1+x{)}^n$ are

${}^n{C}_{r-1}$, ${}^n{C}_r$, ${}^n{C}_{r+1}$

By condition, ${}^n{C}_{r-1}$ : ${}^n{C}_r$ : ${}^n{C}_{r+1}$ = 6 : 33 : 110

Now ${}^n{C}_{r-1}$ : ${}^n{C}_r$ = 6 : 33

⇒ 2n - 13r + 2 = 0 ..............(i)

and ${}^n{C}_r$ : ${}^n{C}_{r+1}$ = 33: 110

⇒ 3n - 13r - 10 = 0 ...............(ii)

Solving (i) and (ii), we get n = 12 and r = 2.

Aliter: We take first n = 4 [By alternate (a)] but

(a) does not hold. Similarity (b).

So alternate (c), n = 12 gives

$(1+x{)}^{12}$ = [1 + 12x + $\frac{12.11}{2.1}$ $x^2$ + $\frac{\mathrm{12.11.10}}{\mathrm{3.2.1}}$$x^3$ + .........]

So coefficient of II, III and IV terms are

12, 6 × 11, 2 × 11 × 10.

So, 12 : 6 × 11 : 2 × 11 × 10 = 6 : 33 : 110.