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TS EAMCET 2015 Solved Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

If

x=1+\;\frac{3}{1!} \times \;\frac{1}{6}+\;\frac{3 \times 7}{2!}\left(\;\frac{1}{6}\right)^{2}

+\;\frac{3 \times 7 \times 11}{3!}\left(\;\frac{1}{6}\right)^{3}+\dots

x^{4}

equals

$81$
$54$
$27$
$8$

Solution:

\because \therefore x=1+\;\frac{3}{1!} \times \;\frac{1}{6}+\;\frac{3 \times 7}{2!}\left(\;\frac{1}{6}\right)^{2}

+\;\frac{3 \times 7 \times 11}{3!}\left(\;\frac{1}{6}\right)^{3}+\dots

\therefore \; \; (1-\alpha)^{-\frac{p}{q}}

+\;\frac{\;\frac{p}{q}\left(\frac{p}{q}+1\right)\left(\frac{p}{q}+2\right)}{3!}\alpha^{3}+\dots

=1+\;\frac{p}{1!}\left(\;\frac{\alpha}{q}\right)+\;\frac{p(p+q)}{2!}\left(\;\frac{\alpha}{q}\right)^{2}

+\;\frac{p(p+q)(p+2q)}{3!}\left(\;\frac{\alpha}{q}\right)^{3}+\dots

On comparing Eqs. (i) and (ii), we get

p=3,\;p+q=7,\;p+2q=11

and

\;\; \;\frac{\alpha}{q}=\;\frac{1}{6}\Rightarrow q=4

and

\alpha=\;\frac{4}{6}=\;\frac{2}{3}

So, let

X=(1-\alpha)^{-\frac{p}{q}}=\left(1-\;\frac{2}{3}\right)^{-\frac{3}{4}}

=\left(\;\frac{1}{3}\right)^{-\frac{3}{4}}=(3)^{\frac{3}{4}}

\therefore \;\; x^{4}=3^{3}=27

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