Question

Find the middle term in the expansion of $[3-\frac{x^3}{6}]^7$

• A. $$T_4=\frac{105}{8}x^4, T_5=\frac{35}{48}x^{12}$$
• B. $$T_4=\frac{-105}{8}x^3, T_5=\frac{35}{48}x^{2}$$
• C. $$T_4=\frac{-105}{8}x^4, T_5=\frac{35}{48}x^{12}$$
• D. $T_4=\frac{-105}{8}x^2, T_5=\frac{35}{48}x^{2}$

Answer 1

Answer:

$$T_4=\frac{-105}{8}x^4, T_5=\frac{35}{48}x^{12}$$

$\frac{(n+1)}{2}^{th};(\frac{n+1}{2}+1)^{th}$

$\frac{7+1}{2}=\frac{8}{2}=4^{th} ;( 4+1)^{th}=5^{th}$

$T_{r+1}=^nC_ra^{n-r}b^r$

$T_4=^7C_33^4(\frac{-x^3}{6})^3$

$=\frac{7!}{3!4!}3^4 \times \frac{(-x^3)^3}{6^3}$

$=\frac{7 \times 6 \times 5 \times 4!}{3! \times 4!} \times \frac{3 \times 3 \times 3 \times 3}{6 \times6 \times 6 } - x^9$

$=\frac{-105}{8}x^4$

$T_4=\frac{-105}{8}x^4$

a=3

b=\frac{-x^3}{6}

n=7

r+1=5

r=4

n-r=3

=\frac{7 \times 5}x^{12}{8 \times 6}=\frac{35}{48}x^{12}

$T_5=\frac{35}{48}x^{12}$