Find the term independent of $x$ in the expansion of ${(x^{2} + \frac{1}{x})}^{9}$

Open in App

Solution

$\begin{matrix} {(x^{2} + \frac{1}{x})}^{9} a c c o r d i n g t o q u e s t i o n, \Rightarrow T_{r + 1} =^{9} C_{r} {(\frac{1}{x})}^{r} {(x^{2})}^{9 - r} =^{9} C_{r} x^{- r} x^{18 - 2 r} =^{9} C_{r} x^{18 - 3 r} f o r r = 6 \Rightarrow 18 - 3 r = 0 =^{9} C_{6} \end{matrix}$
Hence, term independent from $x$ is $7 t h$ term which is $^{9} C_{6} = 84$

Suggest Corrections

Similar questions

x

{(x^{2} + \frac{1}{x})}^{9}

x^{9}

(x^{2} - 1 / 3 x)^{9}

x

{(1 - x)}^{2}

{(x + \frac{1}{x})}^{10}

{(x^{2} - \frac{1}{x})}^{6}

x

{(x^{2} - \frac{1}{x})}^{9}

Join BYJU'S Learning Program