Term independent of ' x ' in expansion of left(frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-frac{x-1}{x-

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7.Binomial Theorem

hard

The term independent of ' $x$ ' in the expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$, where $x \neq 0,1$ is equal to $.....$

$110$

$210$

$300$

$400$

(JEE MAIN-2021)

Solution

$\left(\left(x^{1 / 3}+1\right)-\left(\frac{x^{1 / 2}+1}{x^{1 / 2}}\right)\right)^{10}$

$=\left(x^{1 / 3} \frac{1}{x^{1 / 2}}\right)^{10}$

Now General Term

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{r}\left(\mathrm{x}^{1 / 3}\right)^{10-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{x}^{1 / 2}}\right)^{\mathrm{r}}$

For independent term

$\frac{10-r}{3}-\frac{r}{2}=0 \Rightarrow r=4$

$\Rightarrow T_{5}={ }^{10} C_{4}=210$

Standard 11

Mathematics

The coefficient of ${t^{24}}$ in the expansion of ${(1 + {t^2})^{12}}(1 + {t^{12}})\,(1 + {t^{24}})$ is

hard

(IIT-2003)

Suppose $A$ and $B$ are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 n-1}$. If $2 A=5 B$, then $n$ is equal to:

medium

(JEE MAIN-2025)

The term independent of $x$ in the expansion of ${\left( {\frac{1}{2}{x^{1/3}} + {x^{ – 1/5}}} \right)^8}$ will be

easy

Coefficient of $x^6$ in the binomial expansion ${\left( {\frac{{4{x^2}}}{3}\; – \;\frac{3}{{2x}}} \right)^9}$ is

normal

The value of $x$, for which the 6th term in the expansion of ${\left\{ {{2^{{{\log }_2}\sqrt {({9^{x – 1}} + 7)} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x – 1}} + 1)}}}}} \right\}^7}$ is $84$, is equal to

hard

The term independent of ' $x$ ' in the expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$, where $x \neq 0,1$ is equal to $.....$

Solution

Similar Questions

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