JEE Main 2024 · Mathematics · Binomial Theorem · medium

For x=1, the numerically greatest term in the expansion of (2+3x)^12 is the k^th term. The value of k is

Single correct option

For

x=1

, the numerically greatest term in the expansion of

(2+3x)^{12}

is the

k^{\text{th}}

term. The value of

k

Show answer and solution

Answer: (B) $8$

Write

T_{r+1}=\binom{12}{r}2^{12-r}(3x)^r

. For

x=1

the ratio of consecutive terms is

\dfrac{T_{r+1}}{T_r}=\dfrac{13-r}{r}\cdot\dfrac32

. The terms keep increasing while this ratio exceeds

1

\dfrac{3(13-r)}{2r}>1\iff39-3r>2r\iff r<7.8

. So

T_{r+1}>T_r

for

r\le7

, i.e. the terms increase up to

T_8

, and

T_9<T_8

because the ratio at

r=8

\dfrac{3\cdot5}{16}<1

. Hence the greatest term is the

8^{\text{th}}

and

k=8

. (Numerically

T_8=55427328

and

T_7=51963120

.) The choice

k=9

would come from reading

r<7.8

r+1=9

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