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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

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 = $

Similar Questions

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then

The expression $\frac{1}{\sqrt{5 + 4x}}$ can be expanded by the binomial theorem,if

If $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ to $\infty$,then

If $n$ is a positive integer and the coefficient of $x^{10}$ in the expansion of $(1+x)^{15}$ is equal to the coefficient of $x^5$ in the expansion of $(1-x)^{-n}$,then $n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module