Math, asked by shreyamore045, 9 months ago

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

Answers

Answered by shadowsabers03
2

Question:-

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

Solution:-

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

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

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

To get the coefficient of the power of \displaystyle\sf{x^{-15}} in the term should be equated to -15. Thus,

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

\displaystyle\longrightarrow\sf{r=7}

Hence the coefficient is,

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

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

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