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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Section: Mathematics

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Question : 18 of 160

Marks: +1, -0

If

x = \frac{3}{4 \cdot 8} + \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} + \dots,

2x^2+5x=

$\frac{7}{8}$
7
$\frac{7}{16}$
$\frac{7}{4}$

Solution:

Let,

(1+y)^n = 1 + n y + \frac{n(n-1)}{2!} y^2

+ \frac{n(n-1)(n-2)}{3!} y^3 + \dots

Compare the up to third term of given series with above series,

\frac{n(n-1)}{2!} y^2 = \frac{3}{4 \cdot 8}

and

\frac{n(n-1)(n-2)}{3!} y^3 = \frac{3 \cdot 5}{4 \cdot 8 \cdot 12}

And

\frac{n(n-1)(n-2)(n-3)}{4!} y^4

= \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}

By solving the equations,

x = (1+y)^n - 1 - n y

= \left(1 - \frac{1}{2}\right)^{-1/2} - 1 - \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)

\sqrt{2} - \frac{5}{4}

Now,

2x^2+5x = 2\left[2 + \frac{25}{16} - \frac{5}{\sqrt{2}}\right]

+ 5\sqrt{2} - \frac{25}{4}

= 4 + \frac{25}{8} - 5\sqrt{2} + 5\sqrt{2} - \frac{25}{4}

= \frac{7}{8}

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