Question

M=\(\frac{16}{1.5}+\frac{16}{5.9}+....+\frac{16}{2017.2021}\)      ; N =\(\frac{1}{1.7}+\frac{1}{7.13}+....+\frac{1}{2007.2013}\)

Answer 1

\(M=\frac{16}{1.5}+\frac{16}{5.9}+........+\frac{16}{2017.2021}\)

\(M=4.\left(\frac{4}{1.5}+\frac{4}{5.9}+.......+\frac{4}{2017.2021}\right)\)

\(M=4.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.........+\frac{1}{2017}-\frac{1}{2021}\right)\)

\(M=4.\left(1-\frac{1}{2021}\right)\)

\(M=4.\frac{2020}{2021}\)

\(M=\frac{8080}{2021}\)

\(N=\frac{1}{1.7}+\frac{1}{7.13}+.......+\frac{1}{2007.2013}\)

\(N=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+........+\frac{6}{2007.2013}\right)\)

\(N=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+......+\frac{1}{2007}-\frac{1}{2013}\right)\)

\(N=\frac{1}{6}.\left(1-\frac{1}{2013}\right)\)

\(N=\frac{1}{6}.\frac{2012}{2013}\)

\(N=\frac{1006}{6039}\)

Answer 2

\(N=\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{2007.2013}\)

\(N=\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2007}-\frac{1}{2013}\)

\(N=1-\frac{1}{2013}\)

\(N=\frac{2012}{2013}\)