Question

Find coefficient of x^-^1^5 in the expansion of (3x^2-2/3x^3)^1^0 solve by binomial theorem

Answer 1

Question:-

Find the coefficient of $\sf{x^{-15}}$ in the expansion of $\sf{\left (3x^2-\dfrac {2}{3x^3}\right)^{10}.}$

Solution:-

The $\sf{(r+1)^{th}}$ term of the expansion is,

$\longrightarrow\sf{T_{r+1}=(-1)^r\ ^{10}C_r\ (3x^2)^{10-r}\ \left (\dfrac {2}{3x^3}\right)^r}$

$\longrightarrow\sf{T_{r+1}=(-1)^r\ ^{10}C_r\ 2^r\ 3^{10-2r}\ x^{20-5r}}$

To get the coefficient of $\sf{x^{-15},}$ the power of $\sf{x}$ in the $\sf{(r+1)^{th}}$ term should be equated to -15. Thus,

$\longrightarrow\sf{20-5r=-15}$

$\longrightarrow\sf{r=7}$

Hence the coefficient is,

$\longrightarrow\sf{C\left(T_8\right)=(-1)^7\ ^{10}C_7\ 2^7\ 3^{-4}}$

$\longrightarrow\sf{\underline {\underline {C\left (T_8\right)=-\dfrac {5120}{27}}}}$

Answer 2

$\huge\bold{Question}$

Find the coefficient of $\sf{x^{-15}}$ in the expansion of $\sf{\left (3x^2-\dfrac {2}{3x^3}\right)^{10}.}$

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

The $\sf{(r+1)^{th}}$ term of the expansion is,

$\longrightarrow\sf{T_{r+1}=(-1)^r\ ^{10}C_r\ (3x^2)^{10-r}\ \left (\dfrac {2}{3x^3}\right)^r}$

$\longrightarrow\sf{T_{r+1}=(-1)^r\ ^{10}C_r\ 2^r\ 3^{10-2r}\ x^{20-5r}}$

To get the coefficient of $\sf{x^{-15},}$ the power of $\sf{x}$ in the $\sf{(r+1)^{th}}$ term should be equated to -15. Thus,

$\longrightarrow\sf{20-5r=-15}$

$\longrightarrow\sf{r=7}$

Hence the coefficient is,

$\longrightarrow\sf{C\left(T_8\right)=(-1)^7\ ^{10}C_7\ 2^7\ 3^{-4}}$

$\longrightarrow\sf{\underline {\underline {C\left (T_8\right)=-\dfrac {5120}{27}}}}$