If x = 3/10+3·7/10·15+3·7·9/10·15·20 + ... , then 5x + 8 =

Correct Answer is : 25533{25{5}}{3{3}}33255

Correct Answer is : 25533{25{5}}{3{3}}33255

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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

If x =

\frac{3}{10}+\frac{3\cdot7}{10\cdot15}+\frac{3\cdot7\cdot9}{10\cdot15\cdot20}

+ ... , then 5x + 8 =

Solution:

It is given that

x = \; \frac{3}{10} + \; \frac{3 \cdot 7}{10 \cdot 15} + \; \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \dots

\text{So}

x = \; \frac{3 \cdot 5}{5 \cdot 10} + \; \frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15} + \; \frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20} + \dots

The general term is

T_{r+1} \;\; = \; \frac{3 \cdot 5 \cdot 7 \dots (2r+1)}{5^{r} \cdot 1 \cdot 2 \cdot 3 \dots r}

= \left( \frac{2}{5} \right)^{r} \; \frac{ \; \frac{3}{2} \cdot \frac{5}{2} \cdot \frac{7}{2} \dots \left(r+\frac{1}{2}\right) }{r!}

= \frac{ \left(-\frac{3}{2}\right) \left(-\frac{3}{2}-1\right) \left(-\frac{3}{2}-2\right) \dots \left(-\frac{3}{2}-r+1\right) \left(-\frac{2}{5}\right)^{r} }{r!}

Compare above with general terms of expansion

(1+x)^{n}

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

This gives,

n = -\frac{3}{2}, x = -\frac{2}{5}

Then,

x + \frac{8}{5} \;\; = \left( 1 - \frac{2}{5} \right)^{-\frac{3}{2}}

\;\; = \left( \frac{3}{5} \right)^{-\frac{3}{2}}

\;\; = \left( \frac{5}{3} \right)^{\frac{3}{2}}

\;\; = \frac{5 \sqrt{5}}{3 \sqrt{3}}

So,

\frac{5x+8}{5} \;\; = \frac{5\sqrt{5}}{3\sqrt{3}}

5x+8 \;\; = \frac{25\sqrt{5}}{3\sqrt{3}}

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