If the term independent of x in the expansion of 32x2-13x 9 is k- then 18 k is equal to-

T_r+1= ^9C_r(32x^2 )^9 r( 13x )^r = ^9C_r3^9 2r2^9 r( 1 )^r· x^18 3r For the term independent of x nbsp; ⇒ 18 3r=0 ∴ r=6 ⇒ T_7= ^9C_6(3^ 32^3 ) ∴ k=718 ⇒ 18 ...

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Byju's Answer

Standard XII

Mathematics

Coefficients of Terms Equidistant from Beginning and End

If the term i...

Question

If the term independent of $x$ in the expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $k$ , then $18 k$ is equal to:

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Solution

The correct option is C $7$
$T_{r + 1} =^{9} C_{r} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r}$
$=^{9} C_{r} \frac{3^{9 - 2 r}}{2^{9 - r}} {(- 1)}^{r} \cdot x^{18 - 3 r}$
For the term independent of $x$
$\Rightarrow 18 - 3 r = 0$
$∴ r = 6$
$\Rightarrow T_{7} =^{9} C_{6} (\frac{3^{- 3}}{2^{3}})$
$∴ k = \frac{7}{18}$
$\Rightarrow 18 k = 7$

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If the term independent of x in the expansion of (32x2−13x)9 is k, then 18k is equal to:

The correct option is C 7Tr+1= 9Cr(32x2)9−r(−13x)r = 9Cr39−2r29−r(−1)r⋅x18−3r For the term independent of x ⇒18−3r=0 ∴r=6 ⇒T7= 9C6(3−323) ∴k=718 ⇒18k=7

If the term independent of $x$ in the expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $k$ , then $18 k$ is equal to: