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EAMCET Engineering 2006 Solved Paper

Section: Mathematics

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Question : 89 of 160

Marks: +1, -0

1+\;\frac{2}{4}+\;\frac{2 \cdot 5}{4 \cdot 8}+\;\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\;\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\dots\;\;

$4^{-\frac{2}{3}}$
$\sqrt[3]{16}$
$\sqrt[3]{4}$
$4^{\frac{3}{2}}$

Solution:

Let

S=1+\;\frac{2}{4}+\;\frac{2 \cdot 5}{4 \cdot 8}+\;\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\dots

On comparing with

(1+x)^n=1+nx+\;\frac{n(n-1)}{2!}x^2+\dots,

we get

nx=\;\frac{2}{4}

. . . (i)

and

\;\; \;\frac{n(n-1)}{2!}x^2=\;\frac{2 \cdot 5}{4 \cdot 8}

. . . (ii)

From (i) and (ii)

\;\frac{ \;\frac{n(n-1)}{2!} x^2 }{ n^2 x^2 } = \;\frac{ \;\frac{2 \cdot 5}{4 \cdot 8} }{ \frac{2}{4} \cdot \;\frac{2}{4} }

\Rightarrow \;\; \frac{n-1}{n} = \;\frac{5}{2}

\Rightarrow 2n-2=5n

\Rightarrow n = -\frac{2}{3}

On Putting the value of

n

in Eq. (i), we get

-\frac{2}{3} x = \frac{2}{4}

\Rightarrow x = -\frac{3}{4}

\therefore S=(1+x)^n= \left(1-\frac{3}{4}\right)^{-\frac{2}{3}} = \left(\frac{1}{4}\right)^{-\frac{2}{3}} = \sqrt[3]{16}

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