Question

Let for the 9^th term in the binomial expansion of (3 + 6x)ⁿ, in the increasing powers of 6x, to be the greatest for x = `3/2`, the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k + n₀ is equal to ______.

पर्याय

• 24

• 25

• 26

• 27

Answer 1

Let for the 9^th term in the binomial expansion of (3 + 6x)ⁿ, in the increasing powers of 6x, to be the greatest for x = `3/2`, the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k + n₀ is equal to 24.

Explanation:

Given binomial expression is

(3 + 6x)ⁿ = ⁿC₀3ⁿ + ⁿC₁3^n–1(6x)¹ + ...

General term is shown below,

T_r+1ⁿC_r3^n–r.(6x)r = ⁿC_r3^n–r.6^r.x^r

= `""^n"C"_r3^(n-r).3^r.2^r.(3/2)^r`

= ⁿC_r3ⁿ.3^r  ...`["for"  x = 3/2]`

T₉ is greatest of x = `3/2`

So, T₉ > T₁₀ and T₉ > T₈

Here, 

`"T_9/T_10 > 1` and `T_9/T_8 > 1`

⇒ `(""^nC_8 3^n.3^8)/(""^nC_9 3^n.3^9) > 1` and `(""^nC_8 3^n.3^8)/(""^nC_7 3^n.3^7) > 1`

So, `(""^nC_8)/(""^nC_7) > 1/3` and `(n - 7)/8 > 1/3`

⇒ `29/3 < n < 11`

⇒ n = 10 = n₀

So, in (3 + 6x)ⁿ for n = n₀ = 10

Now, Take (3 + 6x)¹⁰, here T_r+1 = ¹⁰C_r3^10–r6^rx^r

T₇ = ¹⁰C₆3⁴.6⁶.x⁶ = 210.3¹⁰.2⁶x⁶

T₄ = ¹⁰C₃3⁷.6³.x³ = 120.3¹⁰.2³x³

Ratio of coefficient of x⁶ and coefficient of x³ = k

∴ k = `(210.3.^10 2^6)/(120.3^10. 2^3) = 7/4 xx 2^3` = 14

Therefore, k + n₀ = 14 + 10 = 24.