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Army Group X Solved Paper

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Question : 35 of 50

Marks: +1, -0

Find the coefficient of

x^{2}

in the

405
7290
2430
1215

Solution:

Let

x^{2}

contain in

(r+1)

th term in

\;\text{expansion of}\;\left(3x-\frac{1}{x}\right)^{6}

\therefore T_{r+1}={}^{n}C_{r}x^{n-r}a^{r}

=\;\;{}^{5}C_{r}(3x)^{6-r}\left(-\frac{1}{x}\right)^{r}

\;\;={}^{6}C_{r}(3)^{6-r}(-1)^{r}\cdot x^{6-r-r}

\;\;={}^{6}C_{r}(-1)^{r}(3)^{6-r}(x)^{6-2r}

According to the question,

6-2r=2

\Rightarrow 2r=4\Rightarrow r=2

\therefore

Coefficient of

x^{2}

in expansion of

\left(3x-\frac{1}{x}\right)^{6}

={}^{6}C_{2}(-1)^{2}(3)^{6-2}=\frac{6!}{2!4!}(3)^{4}

=\;\frac{6\cdot5}{2}\times81=15\times81=1215

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