Question

The middle term in the expansion of $(3a+\frac{1}{3}{)}^8$ is 1120,where a is a real number.

Find the value of a.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B

2

In the expansion of $(a+b{)}^n$,If n is even, there is only one middle term, which is $(\frac{n+2}{2}{)}^{th}$ term. In this case n = 8.

The middle term is $(\frac{8+2}{2}{)}^{th}$ term or $5^{th}$ term

$T_5=T_{4+1}={}^8{C}_4\left(3a{)}^4\right(\frac{1}{3}{)}^4$

= ${}^8{C}_4(a{)}^4$

It is given ${}^8{C}_4(a{)}^4=1120$

$\frac{8!}{4!\times 4!}(a{)}^4=1120$

$\frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}(a{)}^4=1120$

$70(a{)}^4=1120$

$(a{)}^4=\frac{1120}{70}$

$(a{)}^4=16$

$\Rightarrow$ a = 2