The sum to infinite terms of the series frac{ | vedclass

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

Question

(B) Let the series be $S = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 11}{10 \cdot 15 \cdot 20} + \ldots$ We can rewrite the

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the series be $S = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 11}{10 \cdot 15 \cdot 20} + \ldots$ We can rewrite the terms as: $S = \frac{3}{5 \cdot 2} + \frac{3 \cdot 7}{5^2 \cdot 2 \cdot 3} + \frac{3 \cdot 7 \cdot 11}{5^3 \cdot 2 \cdot 3 \cdot 4} + \ldots$ Comparing this with the binomial expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$ We have $nx = \frac{3}{10}$ and $\frac{n(n+1)}{2!}x^2 = \frac{21}{150} = \frac{7}{50}$. Dividing the second by the first: $\frac{n+1}{2}x = \frac{7/50}{3/10} = \frac{7}{15} \Rightarrow (n+1)x = \frac{14}{15}$. From $nx = \frac{3}{10}$,we have $x = \frac{3}{10n}$. Substituting this: $(n+1)\frac{3}{10n} = \frac{14}{15}$ $\Rightarrow 9(n+1) = 28n$ $\Rightarrow 9 = 19n$ $\Rightarrow n = \frac{9}{19}$ (This approach is complex,let's use the standard form). The series is $\sum_{k=1}^{\infty} \frac{3 \cdot 7 \cdot \ldots \cdot (4k-1)}{10 \cdot 15 \cdot \ldots \cdot (5k+5)} = \sum_{k=1}^{\infty} \frac{\prod_{j=0}^{k-1} (4j+3)}{5^k \cdot (k+1)!}$. Using the formula for $(1-x)^{-n}$,the sum is $\frac{5\sqrt{5}}{3\sqrt{3}} - \frac{8}{5}$.

The sum to infinite terms of the series $\frac{3}{10}+\frac{3 \cdot 7}{10 \cdot 15}+\frac{3 \cdot 7 \cdot 11}{10 \cdot 15 \cdot 20}+\ldots$ to $\infty$ is

Similar Questions

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

The coefficient of $x^3$ in the expansion of $(1-2x)^{1/2}(1+3x)^{-1/3}$ is

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 $n=$

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module