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

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

1. A

2. B

3. C

4. D

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### Solution:

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

$\begin{array}{l}{a}_{k}=\frac{\left(1\right)\left(3\right)\left(5\right)\cdots \left(2k-1\right)}{\left(2\right)\left(4\right)\left(6\right)\cdots \left(2k\right)\left(2k+2\right)}\\ =\frac{\left(1\right)\left(3\right)\left(5\right)\cdots \left(2k-1\right)\left(2k+2-2k-1\right)}{\left(2\right)\left(4\right)\left(6\right)\cdots \left(2k\right)\left(2k+2\right)}\\ ={b}_{k}-{b}_{k+1}\end{array}$

where

Thus,

$\sum _{k=1}^{20} {a}_{k}=\sum _{k=1}^{20} \left({b}_{k}-{b}_{k+1}\right)={b}_{1}-{b}_{21}$