Number of integral tems in expansion of left{7^{left(frac{1}{2}right)}+11^{left(frac{1}{6}righ

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7.Binomial Theorem

hard

Number of integral tems in the expansion of $\left\{7^{\left(\frac{1}{2}\right)}+11^{\left(\frac{1}{6}\right)}\right\}^{824}$ is equal to..................

$142$

$138$

$421$

$456$

(JEE MAIN-2024)

Solution

General term in expansion of $\left((7)^{1 / 2}+(11)^{1 / 6}\right)^{824}$ is $t_{I+1}={ }^{824} C_{\mathrm{r}}(7)^{\frac{824-\mathrm{I}}{2}}(11)^{\mathrm{I} / 6}$

For integral term, $r$ must be multiple of $6$ .

Hence $r=0,6,12, \ldots \ldots \ldots…….. .822$

Standard 11

Mathematics

The coefficient of ${x^7}$ in the expansion of ${\left( {\frac{{{x^2}}}{2} – \frac{2}{x}} \right)^8}$ is

easy

If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, are equal, then the value of $b$ is equal to:

hard

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medium

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medium

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The coefficient of $x^{10}$ in the expansion of $(1 + x)^2 (1 + x^2)^3 ( 1 + x^3)^4$ is euqal to

hard

(JEE MAIN-2018)

Number of integral tems in the expansion of $\left\{7^{\left(\frac{1}{2}\right)}+11^{\left(\frac{1}{6}\right)}\right\}^{824}$ is equal to..................

Solution

Similar Questions

The coefficient of ${x^7}$ in the expansion of ${\left( {\frac{{{x^2}}}{2} – \frac{2}{x}} \right)^8}$ is

If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, are equal, then the value of $b$ is equal to:

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 …… .

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