Question

Find the term independent of x in the expansion of (3x^2/2-1/3x)^9

Answer 1

In the expansion:

( a + b )^n :  a = 3 x² / 2,    b = 1 / 3 x ,     n = 9

T (k+1) = n C k · a^(n-k )· b^k

When the term is independent of x:

( x² )^(9-k) · (1/x)^k = 1

x^(18-2k)· (x)^(-k) = 1

x^(18-3k) = 1

18-3k = 0

3k = 18

k = 18 : 3 = 6

T(k+1) = T(6+1) = T7

Answer: This is the 7th term.

Answer 2

Answer: $\dfrac{84}{216}$ which is seventh term

Step-by-step explanation:

To find the term independent of x in the expansion $(\dfrac{3x^2}{2} - \dfrac{1}{3x} )^9$

The $n^{th}$ term of the expansion is $(a+b)^n= _{r}^{n}\textrm{C} \times a^{(n-r)} b^{r}$

Therefore

$(\dfrac{3x^2}{2} - \dfrac{1}{3x} )^9=_{r}^{9}\textrm{C} (\dfrac{3x^2}{2})^{9-r} ( \dfrac{1}{3x})^r\\\\= \dfrac{9!}{r!(9-r)!} \times (\dfrac{3}{2}x^2) ^{9-r} \dfrac{1}{3^rx^r} \\\\ \text {therefore for independent of x put n= 6}\\\\ \dfrac{9!}{6!(9-6)!} \times (\dfrac{3}{2}x^2) ^{9-6} \dfrac{1}{3^6x^6}\\\\\\ 84 \times \dfrac{3^3}{8} x^6 \times \dfrac{1}{3^6\times x^6} =\dfrac{84}{8\times 3^3} =\dfrac{84}{216}$

Hence, seventh term  $T_{6+1}$ = $T_{7}$ is independent of x