Term independent of x in expansion of left(√[3]{x}+frac{1}{2 √[3]{x}}right)^{18}, x,>,0

English

Gujarati

Hindi

7.Binomial Theorem

medium

Find the term independent of $x$ in the expansion of $\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x\,>\,0$

Option A

Option B

Option C

Option D

Solution

We have ${T_{r + 1}} = {\,^{18}}{C_r}{(\sqrt[3]{x})^{18 – r}}{\left( {\frac{1}{{2\sqrt[3]{x}}}} \right)^r}$

$ = {\,^{18}}{C_r}{x^{\frac{{18 – r}}{3}}} \cdot \frac{1}{{{2^r} \cdot {x^{\frac{r}{3}}}}} = {\,^{18}}{C_r}\frac{1}{{{2^r}}} \cdot {x^{\frac{{18 – 2r}}{3}}}$

Since we have to find a term independent of $x$, i.e., term not having $x$, so take $\frac{18-2 r}{3}=0$

We get $r=9 .$ The required term is ${\,^{18}}{C_9}\frac{1}{{{2^9}}}$

Standard 11

Mathematics

If the coefficients of $a^{r-1}, a^{r}$ and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression, prove that $n^{2}-n(4 r+1)+4 r^{2}-2=0$

hard

Write the general term in the expansion of $\left(x^{2}-y x\right)^{12}, x \neq 0$

easy

Coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is

medium

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\ldots+x^{10}$ is

hard

(JEE MAIN-2020)

Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\right)^{n}, x \neq 0,$ be in the ratio $12: 8: 3 .$ Then the term independent of $x$ in the expansion, is equal to …… .

medium

(JEE MAIN-2021)

Find the term independent of $x$ in the expansion of $\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x\,>\,0$

Solution

Similar Questions

If the coefficients of $a^{r-1}, a^{r}$ and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression, prove that $n^{2}-n(4 r+1)+4 r^{2}-2=0$

Write the general term in the expansion of $\left(x^{2}-y x\right)^{12}, x \neq 0$

Coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\ldots+x^{10}$ is

Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\right)^{n}, x \neq 0,$ be in the ratio $12: 8: 3 .$ Then the term independent of $x$ in the expansion, is equal to …… .

Start a Free Trial Now