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AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

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Question : 18 of 160

Marks: +1, -0

If

\alpha = \frac{5}{2! \times 3} + \frac{5 \times 7}{3! \times 3^3} + \frac{5 \times 7 \times 9}{4! \times 3^3} + \ldots,

\alpha^2 + 4\alpha =

21
23
25
27

Solution:

Consider the expression,

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

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

+ \frac{n(n-1)(n-2)(n-3)}{4!} x^4 + \ldots)

Comparison the given equation with above equation,

\frac{n(n-1)}{2!} x62 = \frac{5}{2! \times 3}

and

\frac{n(n-1)(n-2)}{3!} x^3 = \frac{5(7)}{3!(3^2)}

So,

x = -\frac{2}{3}

and

n = -\frac{3}{2}

Therefore,

\left(1 - \frac{2}{3}\right)^{-\frac{3}{2}} = 1 + 1 + \alpha

3^{3/2} = 2 + \alpha

\alpha^2 + 4\alpha + 4 = 27

\alpha^2 + 4\alpha = 23

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