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If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+kx)^9$ are equal then the value of $k$ is

$\begin{array}{1 1}(A)\;-\large\frac{9}{7}\\(B)\;\large\frac{9}{7}\\(C)\;\large\frac{7}{9}\\(D)\;\text{None of these}\end{array} $

1 Answer

  • $T_{r+1}=nC_ra^{n-r} b^r$
$T_{r+1}$ in $(3+kx)^9=9C_r 3^{9-r}(kx)^r$
$\Rightarrow 9C_r 3^{9-r}k^r x^r$
$\therefore $ Coefficient of $x^r=9C_r 3^{9-r} k^r$
Now coefficient of $x^2$=Coefficient of $x^3$
$\therefore 9C_23^{9-2}k^2=9C_33^{9-3}k^3$
$\Rightarrow 36\times 3^7 k^2=84\times 3^6k^3$
$\Rightarrow 36=28k$
$\Rightarrow k=\large\frac{9}{7}$
Hence (B) is the correct answer.
answered Jun 26, 2014 by sreemathi.v

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