Coefficient of {x^{ - 9}} in expansion of {left( {frac{{{x^2}}}{2}

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7.Binomial Theorem

easy

The coefficient of ${x^{ - 9}}$ in the expansion of ${\left( {\frac{{{x^2}}}{2} - \frac{2}{x}} \right)^9}$ is

$512$

$-512$

$521$

$251$

Solution

(b) Here ${T_{r + 1}} = {}^{\rm{9}}{C_r}{\left( {\frac{{{x^2}}}{2}} \right)^{9 – r}}{\left( {\frac{{ – 2}}{x}} \right)^r}$

= ${}^9{C_r}\frac{{{x^{18 – 3r}}{{( – 2)}^r}}}{{{2^{9 – r}}}},$ this contains ${x^{ – 9}}$ if $18 -3r = -9$ i.e. if $r = 9$. Coefficient of ${x^{ – 9}}$

= ${}^9{C_9}\frac{{{{( – 2)}^9}}}{{{2^0}}} = – {2^9} = – 512$.

Standard 11

Mathematics

Find the $r^{\text {th }}$ term from the end in the expansion of $(x+a)^{n}$

medium

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

medium

(JEE MAIN-2020)

In the expansion of the following expression $1 + (1 + x) + {(1 + x)^2} + ….. + {(1 + x)^n}$ the coefficient of ${x^k}(0 \le k \le n)$ is

hard

The term independent of $x$ in the expansion of ${\left( {{x^2} – \frac{{3\sqrt 3 }}{{{x^3}}}} \right)^{10}}$ is

easy

If the coefficient of the second, third and fourth terms in the expansion of ${(1 + x)^n}$ are in $A.P.$, then $n$ is equal to

medium

(IIT-1994)

The coefficient of ${x^{ - 9}}$ in the expansion of ${\left( {\frac{{{x^2}}}{2} - \frac{2}{x}} \right)^9}$ is

Solution

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