In the expansion of \(\Big(\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9},\) the term without x is equal to:

In the expansion of \(\Big(\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9},\) the term without x is equal to:

  1. None of these.

  2. \(\frac{28}{243}\)

  3. \(\frac{28}{81}\)

  4. \(\frac{-28}{243}\)

Solution

Solution:

Suppose the (r + 1)th term in the given expansion is independent of x.

Then, we have

\(\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}(\text{x}^{2})^{9-\text{r}}\Big(\frac{-1}{3\text{x}}\Big)^{\text{r}}\)

\(=(-1)^{\text{r}}\ {^\text{9}}\text{C}_{\text{r}}\frac{1}{3\text{r}}\ \text{x}^{18-2\text{r}-\text{r}}\)

For this term to be independent of x, we must have

\(18-3\text{r}=0\)

\(\Rightarrow \text{r}=6\) 

\(\therefore\) Required term \(=(-1)^{6}\ {^\text{9}}\text{C}_{\text{6}}\ \frac{1}{3^{6}}=\frac{9\times8\times7}{3\times2}\times\frac{1}{3^{6}}=\frac{28}{243}\)

Asked in: RDSHARMA

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