The sum of 1 + frac{2}{1 times 2 times 3} + f | vedclass

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

Question

(A) The general term of the series after the first term $1$ is $T_n = \frac{2}{(2n-1)(2n)(2n+1)}$ for $n \ge 1$.Using partial fractions,$\frac{2}{(2n-

Vedclass · Accepted Answer

$2 \ln 2$

Vedclass · Answer

(A) The general term of the series after the first term $1$ is $T_n = \frac{2}{(2n-1)(2n)(2n+1)}$ for $n \ge 1$.Using partial fractions,$\frac{2}{(2n-1)(2n)(2n+1)} = \frac{1}{(2n-1)(2n)} - \frac{1}{2n(2n+1)} = \left( \frac{1}{2n-1} - \frac{1}{2n} ight) - \left( \frac{1}{2n} - \frac{1}{2n+1} ight) = \frac{1}{2n-1} - \frac{2}{2n} + \frac{1}{2n+1}$.Sum $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{2}{2n} + \frac{1}{2n+1} ight)$.Expanding the series: $S = 1 + (1 - 1 + 1/3) + (1/3 - 1/2 + 1/5) + (1/5 - 1/3 + 1/7) + \dots$Alternatively,recognize the series as $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} ight) - \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n+1} ight)$.This simplifies to $2 	imes (1 - 1/2 + 1/3 - 1/4 + \dots) = 2 \ln 2$.

The sum of $1 + \frac{2}{1 \times 2 \times 3} + \frac{2}{3 \times 4 \times 5} + \frac{2}{5 \times 6 \times 7} + \dots$ is

Similar Questions

$\cosh^{-1} 2 = $

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 $y = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{x^4}{4!} + \dots$,then $x = $

The sum to infinity of the given series $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ is

If $P = 1 + \frac{1}{2 \times 2} + \frac{1}{3 \times 2^{2}} + \dots$ and $Q = \frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module