The sum of the series \; 3/4 · 8 - 3 · 5/4 · 8 · 12 + 3 · 5 · 7/4 · 8 · 12 · 16 -

Correct Answer is : 23−34{2/3} - 3/432−43

Correct Answer is : 23−34{2/3} - 3/432−43

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EAMCET Engineering 2007 Solved Paper

Section: Mathematics

Question : 90 of 160

Marks: +1, -0

The sum of the series

\; \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

Solution:

Given series is

\;\; \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 + \frac{3}{4} - \frac{3}{4}

= \;\; \frac{3}{4} + \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 - \frac{3}{4}

= \; 1 - \frac{1}{4} + \frac{1 \cdot 3}{2 \cdot 4 \cdot 4} - \frac{1 \cdot 3 \cdot 5}{4 \cdot 4 \cdot 2 \cdot 4 \cdot 3} + \dots - \frac{3}{4}

= \; \left[ 1 + 1 \left(-\frac{1}{4}\right) + \frac{1(1+2)}{2} \left(-\frac{1}{4}\right)^2. \right. \text{(unclosed bracket)}

\; + \left[\frac{1(1+2)(1+4)}{3 \cdot 2} \left(-\frac{1}{4}\right)^3 + \dots \right] - \frac{3}{4}

\; = \left[ 1 + \frac{1}{1!} \left(-\frac{1}{4}\right) + \frac{1(1+2)}{2!} \left(-\frac{1}{4}\right)^2 + . \right. \text{(unclosed bracket)}

\; \left[ \frac{1(1+2)(1+4)}{3!} \left(-\frac{1}{4}\right)^3 + \dots \right] - \frac{3}{4}

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

\; = \sqrt{\frac{2}{3}} - \frac{3}{4}

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