Question

find out term independent of X in the expansion of (9x2-1*3x)12​

Answer 1

Answer:

answer is 495

Explanation:

mark me as brainliest

Answer 2

Note: The correct question must be-

Find out term independent of $x$ in the expansion of $(9x^{2} -\frac{1}{3x} )^{12}$.

Step 1: Given data

Given expression, $(9x^{2} -\frac{1}{3x} )^{12}$

Term independend of $x=?$

Step 2: Evaluating the given expression

Assume, that $(r+1)^{th}$ term in the expansion is $x$ free.

We know, for $(a+b)^{n}$, $t_{r+1} =_{}^{n}\textrm{C}_{r}a^{n-r}b^{r}$

Now,

$t_{r+1} =_{}^{12}\textrm{C}_{r}(9x^{2}) ^{12-r}(-\frac{1}{3x}) ^{r}$

$t_{r+1} =_{}^{12}\textrm{C}_{r}9^{12-r}x ^{24-2r}(-\frac{1}{3x}) ^{r}$

$t_{r+1} =_{}^{12}\textrm{C}_{r}9^{12-r}x ^{24-2r}(-\frac{1}{3}) ^{r}x^{-r}$

$t_{r+1} =_{}^{12}\textrm{C}_{r}9^{12-r}x ^{24-2r-r}(-\frac{1}{3}) ^{r}$

$t_{r+1} =_{}^{12}\textrm{C}_{r}9^{12-r}x ^{24-3r}(-\frac{1}{3}) ^{r}$

Now, since $t_{r+1}\ is\ independent \ of\ x$

So, index of $x$ will be zero

Thus,

$24-3r=0\\3r=24\\r=8$

$t_{r+1}=t_{8+1}= t_{9}=_{}^{12}\textrm{C}_{8}9^{12-8}x ^{0}(-\frac{1}{3}) ^{8}$

$t_{9}= _{}^{12}\textrm{C}_{8}9^{4}(\frac{1}{3^{8}})\\\\t_{9}= _{}^{12}\textrm{C}_{8}3^{8}(\frac{1}{3^{8}})\\t_{9}= _{}^{12}\textrm{C}_{8}$

Hence, $9^{th}$ term is independent of $x$ which is $_{}^{12}\textrm{C}_{8}$.

#SPJ3