Question

$\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 10}+\dots \dots \dots \dots \dots ...\infty$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (C)

$T_n=\frac{1\cdot 3\cdot 5\dots \dots \dots (2n-1)}{2\cdot 4\cdot 6\dots \dots \dots \dots .2n(2n+2)}[2n+2)-(2n+1)$
$T_n=\frac{\mathrm{1.3.5}\dots \dots \dots (2n-1)}{\mathrm{2.4.6}\dots \dots \dots \dots .2n}-\frac{1\cdot 3\cdot 5\dots \dots \dots (2n-1)(2n+1)}{\mathrm{2.4.6}\dots \dots \dots \dots .2n(2n+2)}$
$\therefore S_n=\sum T_n=\frac{1}{2}-\frac{1\cdot 3\cdot 5\dots \dots \dots (2n+1)}{2\cdot 4\cdot 6\dots \dots \dots ..2n(2n+2)}\text{ Note that }{S}_{\infty }=\frac{1}{2}$