In the expansion of \(\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{\frac{-1}{5}}\Big)^{8},\) the term…

In the expansion of \(\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{\frac{-1}{5}}\Big)^{8},\) the term independent of x is:

  1. \(\text{T}_{6}\)

  2. \(\text{T}_{7}\)

  3. \(\text{T}_{5}\)

  4. \(\text{T}_{8}\)

Solution

Solution:

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

Thus, we have

\(\text{T}_{\text{r}+1}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-\text{r}}\Big(\text{x}^{\frac{-1}{5}}\Big)^{\text{r}}\)

\(={^\text{8}}\text{C}_{\text{r}}\frac{1}{2^{8-\text{r}}}\ \text{x}^{\frac{8-\text{r}}{3}-\frac{\text{r}}{5}}\)

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

\(\frac{8-\text{r}}{3}-\frac{\text{r}}{5}=0\)

\(\Rightarrow 40-5\text{r}-3\text{r}=0\)

\(\Rightarrow \text{r}=5\)

Hence, the required term is the 6th term, i.e. \(\text{T}_{6}\)

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