If x = frac{3}{10} + frac{3 cdot 7}{10 cdot 1 | vedclass

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(D) The given series is $x = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \ldots$ Multiply and divi

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$\frac{25 \sqrt{5}}{3 \sqrt{3}}$

Vedclass · Answer

(D) The given series is $x = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \ldots$ Multiply and divide by $5$ to adjust the terms: $x = \frac{3 \cdot 5}{5 \cdot 10} + \frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15} + \frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20} + \ldots$ The general term $T_r$ for $r \geq 1$ is $T_r = \frac{3 \cdot 5 \cdot 7 \cdots (2r+1)}{5 \cdot 10 \cdot 15 \cdots (5r)} = \frac{3 \cdot 5 \cdot 7 \cdots (2r+1)}{5^r \cdot r!}$. Using the binomial expansion $(1-y)^{-n} = 1 + ny + \frac{n(n+1)}{2!}y^2 + \cdots$,we identify $n = 3/2$ and $y = 2/5$. Thus,$1 + x = (1 - 2/5)^{-3/2} = (3/5)^{-3/2} = (5/3)^{3/2} = \frac{5\sqrt{5}}{3\sqrt{3}}$. Adding $1$ to both sides of the series: $1 + x = 1 + \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \cdots = (1 - 2/5)^{-3/2} = (3/5)^{-3/2} = \frac{5\sqrt{5}}{3\sqrt{3}}$. Therefore,$x = \frac{5\sqrt{5}}{3\sqrt{3}} - 1$. Then $5x + 8 = 5(\frac{5\sqrt{5}}{3\sqrt{3}} - 1) + 8 = \frac{25\sqrt{5}}{3\sqrt{3}} - 5 + 8 = \frac{25\sqrt{5}}{3\sqrt{3}} + 3$.

If $x = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \ldots$,then $5x + 8 = $

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