#### Find the term independent of x in the expansion of [math] (\sqrt[3]{x} + \frac{1}{2\sqrt[3]{x}})^{18} [/math], x 0

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Expanding binomials (video) | Polynomials | Khan Academy

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Integral $\int_{-**1**}^**1**\frac1x\**sqrt**{\**frac**{**1**+**x**}{**1**-**x**}}

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Krishna

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Given expansion

[math] (\sqrt[3]{x} + \frac{1}{2\sqrt[3]{x}})^{18} [/math], x > 0

Find the term independent

We know that

The (r + 1)th term of the expansion (x + y)^n is given by T_{r + 1} = ^nC_r x^{n-r} y^r

[math] (\sqrt[3]{x} + \frac{1}{2\sqrt[3]{x}})^{18} [/math]

n = 18 [math]x\ =\ \sqrt[3]{x}\ \ and\ \ \ y\ =\ \frac{1}{2\sqrt[3]{x}}[/math]

[math] T_{r + 1} = ^{18}C_r (\sqrt[3]{x})^{18 - r} (\frac{1}{2\sqrt[3]{x}})^r [/math]

T_{r+1}=^{18}C_r(x^{\frac{1}{3}})^{18-r}\ \left(\frac{1}{2}\right)^{^r}\ \frac{1}{\left(x^{\frac{1}{3}}\right)^{^r}}

T_{r+1}=^{18}C_r\ \ \ x^{\frac{18\ -\ r}{3}}\ \ \ x^{-\frac{\left\{r\right\}}{3}}\left(\frac{1}{2}\right)^{^r}\ ^{ }

T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -\ r}{3}\ -\ \frac{r}{3}}\ \ \left(\frac{1}{2}\right)^{^r}

T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -\ r-r}{3}\ }\ \ \left(\frac{1}{2}\right)^r

T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -2\ r}{3}\ }\ \ \left(\frac{1}{2}\right)^r.................(1)

Since we have to find a term independent of x, i.e., term not having x,

so take x^{ \frac{18 - 2r}{3}} = x^{0}

\frac{18 - 2r}{3} = 0

18 - 2r = 0

18 = 2r

r = \frac{18}{2}

r = 9

Substituting the r = 9 in equation (1)

T_{r+1}=^{18}C_r\ \ \ \ \ x^{\frac{18\ -2\ r}{3}\ }\ \ \left(\frac{1}{2}\right)^r

T_{r+1}=^{18}C_9\ \ \ \ \ x^{\frac{18\ -2\ \left(9\right)}{3}\ }\ \ \left(\frac{1}{2}\right)^9

T_{r+1}=^{18}C_9\ \ \ \left(\frac{1}{2}\right)^9\ x^0

Hence, required term = ^{18}C_{9\ }\ \frac{1}{2^9}