JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Binomial Theorem

Let for the

9^{th}

term in the binomial expansion of

(3 + 6 x)^{n}

, in the increasing powers of

6 x

, to be the greatest for

x = \frac{3}{2}

, the least value of n is

n_{0}

. If k is the ratio of the coefficient of

x^{6}

to the coefficient of

x^{3}

, then

k + n_{0}

is equal to:

Answer: 24

Solution

For

x = 3/2

, the general term becomes

T_{r + 1} = (r n) 3^{n} \cdot 3^{r}

. For

T_{9}

to be the greatest, we need

T_{9} > T_{8}

and

T_{9} > T_{10}

. This gives

\frac{n - 8}{9} > \frac{1}{3}

and

\frac{n - 7}{8} > \frac{1}{3}

, yielding

\frac{29}{3} < n < 11

. Thus

n_{0} = 10

. In

(3 + 6 x)^{10}

, the coefficient of

x^{6}

(6 10) 3^{4} \cdot 6^{6} = 210 \cdot 3^{10} \cdot 2^{6}

and the coefficient of

x^{3}

(3 10) 3^{7} \cdot 6^{3} = 120 \cdot 3^{10} \cdot 2^{3}

. The ratio

k = \frac{210 \cdot 2 ^{6}}{120 \cdot 2 ^{3}} = 14

. Therefore

k + n_{0} = 24

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer