If the sum of the binomial coefficients of the expansion $\Big(2\text{x}+\frac{1}{\text{x}}\Big)^{\text{n}}$ is equal to 256, then the term independent of x is:

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Aicharya · Accepted Answer

Solution: Suppose (r + 1) th term in the given expansion is independent of x. Then, we have $\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}(2\text{x})^{\text{n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$ $={^\text{n}}\text{C}_{\text{r}}(2)^{\text{n}-\text{r}}\text{x}^{\text{n}-2\text{r}}$ For this term to be independent of x, we must have $\text{n}-2\text{r}=0$ $\Rightarrow \text{r}=\frac{\text{n}}{2}$ $\therefore$ Required term $={^\text{n}}\text{C}_{\frac{\text{n}}{2}}\ 2^{\text{n}-\frac{\text{n}}{2}}=\frac{\text{n!}}{\big[(\frac{\text{n}}{2})\big]}\ 2^{\frac{\text{n}}{2}}$ We know, Sum of the given expansion = 256 Thus, we have $2^{\text{n}}.1^{\text{n}}=256$ $\Rightarrow \text{n}=8$ $\therefore$ Required term $=\frac{8!}{(4)!(4)!}2^{4}=1120$

If the sum of the binomial coefficients of the expansion…

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