The sum to the infinite terms to the series 3/10 + 3 · 7/10 · 15 + 3 · 7 · 9/10 · 15 · 20 + ·s ∞ is

Correct Answer is : 5533−85{5 {5}}{3 {3}} - 8/53355−58

Correct Answer is : 5533−85{5 {5}}{3 {3}} - 8/53355−58

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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

\frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20}

+ \cdots \infty

Solution:

Consider the series,

\frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20}

+ \cdots

= \frac{3}{5 \times 2!} + \frac{3 \cdot 7}{5^{2} \cdot 3!} + \frac{3 \cdot 7 \cdot 9}{5^{3} \cdot 4!} + \cdots

Consider the expression,

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

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

On comparison, we get

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

And,

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

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

........(I)

And,

n(n-1)(n-2) x^{3} = \frac{21}{25}

.......(II)

This implies,

(n-2) x \left( \frac{3}{5} \right) = \frac{21}{25}

x = \frac{7}{5(n-2)}

Substitute the value of x in equation (1).

n(n-1) \frac{49}{25 (n-2)^{2}} = \frac{3}{5}

34 n^{2} + 11 m - 60 = 0

This gives,

n = -\frac{3}{2}

n = \frac{20}{17}

So,

x = \frac{7}{5} \cdot \frac{1}{\left( -\frac{3}{2} - 2 \right)}

= -\frac{2}{5}

The sum of the series is given by,

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

= \frac{5 \sqrt{5}}{3 \sqrt{3}} - \frac{8}{5}

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