The coefficient of the middle term in the expansion of (2 + 3x)4 is :

Question

The coefficient of the middle term in the expansion of (2 + 3x)4 is : (A) 6 (B) 5! (C) 8! (D) 216

Tardigrade Admin · Accepted Answer

When exponent is n then total number of terms are n+1. So, total number of terms in (2+3 x) 4=5 Middle term is 3 rd . ⇒ T3= 4 C2(2)2 ⋅(3 x)2 =(4 × 3 × 2 × 1/2 × 1 × 2) × 4 × 9 x2 =216 x2

Q. The coefficient of the middle term in the expansion of $(2 + 3 x)^{4}$ is :

6

5!

8!

216

When exponent is $n$ then total number of terms are $n + 1$ .
So, total number of terms in $(2 + 3 x) 4 = 5$
Middle term is $3 r d . \Rightarrow T_{3} =^{4} C_{2} (2)^{2} \cdot (3 x)^{2}$
$= \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2} \times 4 \times 9 x^{2}$
$= 216 x^{2}$

Q. The coefficient of the middle term in the expansion of (2+3x)4 is :

6

5!

8!

216

When exponent is n then total number of terms are n+1. So, total number of terms in (2+3x)4=5 Middle term is 3rd.⇒T3​=4C2​(2)2⋅(3x)2 =2×1×24×3×2×1​×4×9x2 =216x2

Q. The coefficient of the middle term in the expansion of $(2 + 3 x)^{4}$ is :

When exponent is $n$ then total number of terms are $n + 1$ .
So, total number of terms in $(2 + 3 x) 4 = 5$
Middle term is $3 r d . \Rightarrow T_{3} =^{4} C_{2} (2)^{2} \cdot (3 x)^{2}$
$= \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2} \times 4 \times 9 x^{2}$
$= 216 x^{2}$