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

$\Rightarrow$ 2n - 13r + 2 = 0 ..............(i)

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

$\Rightarrow$ 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

$12,6\times 11,2\times 11\times 10.$

So, $12:6\times 11:2\times 11\times 10=6:33:110.$