frac{1}{1 cdot 2 cdot 3} + frac{1}{3 cdot 4 c | vedclass

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

Question

(B) The general term of the series is $T_n = \frac{1}{(2n - 1)(2n)(2n + 1)}$.Using partial fractions,we have $T_n = \frac{1}{2} \left[ \frac{1}{2n - 1

Vedclass · Accepted Answer

$\log_{e} 2 - \frac{1}{2}$

Vedclass · Answer

(B) The general term of the series is $T_n = \frac{1}{(2n - 1)(2n)(2n + 1)}$.Using partial fractions,we have $T_n = \frac{1}{2} \left[ \frac{1}{2n - 1} - \frac{2}{2n} + \frac{1}{2n + 1} \right] = \frac{1}{2} \left[ \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) \right]$.Summing from $n=1$ to $\infty$:$S = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right)$.We know the logarithmic series expansion $\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right)$.Also,$\sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = 1 - \log_{e} 2$.Substituting these values:$S = \frac{1}{2} (\log_{e} 2) - \frac{1}{2} (1 - \log_{e} 2) = \frac{1}{2} \log_{e} 2 - \frac{1}{2} + \frac{1}{2} \log_{e} 2 = \log_{e} 2 - \frac{1}{2}$.

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{5 \cdot 6 \cdot 7} + \dots \infty = $

Similar Questions

$\frac{1}{1 \cdot 3} + \frac{1}{2} \cdot \frac{1}{3 \cdot 5} + \frac{1}{3} \cdot \frac{1}{5 \cdot 7} + \dots \infty = $

The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is:

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:

If $\log (1 - x + {x^2}) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \dots$,then ${a_3} + {a_6} + {a_9} + \dots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module