Sum of the first 20 terms of the series 1(2)(4)+(1)(3)(2)(4)(6)+(1)(3)(5)(2)(4)(6)(8)+… is

A
$\frac{1}{2} - \frac{1}{2^{40}} (^{40} C_{20})$
B
$\frac{1}{2} - \frac{1}{2^{41}} (^{42} C_{21})$
C
$\frac{1}{2} - \frac{1}{2^{42}} (^{42} C_{21})$
D
$\frac{1}{2} - \frac{1}{2^{43}} (^{40} C_{20})$

Solution:

Let $a_{k}$ denote the kth term of the series, then

$\begin{array}{l} a_{k} = \frac{(1) (3) (5) \dots (2 k - 1)}{(2) (4) (6) \dots (2 k) (2 k + 2)} \\ = \frac{(1) (3) (5) \dots (2 k - 1) (2 k + 2 - 2 k - 1)}{(2) (4) (6) \dots (2 k) (2 k + 2)} \\ = b_{k} - b_{k + 1} \end{array}$

where

$\begin{array}{l} b_{k} = \frac{(1) (3) (5) \dots (2 k - 1)}{(2) (4) (6) \dots (2 k)} \\ = \frac{(1) (2) (3) (4) \dots (2 k - 1) (2 k)}{[(2) (4) (6) \dots (2 k)]^{2}} \\ = \frac{1}{2^{2 k}} (^{2 k} C_{k}) \end{array}$

Thus,

$\sum_{k = 1}^{20} a_{k} = \sum_{k = 1}^{20} (b_{k} - b_{k + 1}) = b_{1} - b_{21}$

$= \frac{1}{2} - \frac{1}{2^{42}} (^{42} C_{21})$

Sum of the first 20 terms of the series $\frac{1}{(2) (4)} + \frac{(1) (3)}{(2) (4) (6)} + \frac{(1) (3) (5)}{(2) (4) (6) (8)} + \dots$ is

Solution:

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