the sum 1-1/2+1/3-1/4+1/5+ 1/6 up to 1/2013 equals
(A) 1/1006+1/1007+1/1008 up to 1/2013
(B)1/1007+1/1008+1/1009 up to 1/2013
(C) 1/1006+1/1007+1/1008 up to 1/2012
(D)1/1007+1/1008+1/1009 up to 1/2012

Dear Student,

Please find below the solution to the asked query :

$$\begin{gathered} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-.....-\frac{1}{2012}+\frac{1}{2013} \\ = \left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-.....-\frac{1}{2012}+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2012}\right) \\ = \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2012}\right) \\ = \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1006}\right) \\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1006}\right)+\left(\frac{1}{1007}+\frac{1}{1008}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1006}\right) \\ =\left(\frac{1}{1007}+\frac{1}{1008}+.....+\frac{1}{2013}\right) \\ Answer \end{gathered}$$

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