Question

CMR:M=(\(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)).\(\frac{1-3-5-7-9-...-49}{89}=\frac{-9}{28}\)

giúp tớ nhé

Answer 1

Đặt A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)

\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{1}{44.49}\right)\)

\(\Rightarrow\) A = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\)

\(\Rightarrow\)A = \(\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)

Đặt B = \(\frac{1-3-5-7-9-...-49}{89}\)

\(\Rightarrow\)B = \(\frac{1-\left(3+5+7+9+...+49\right)}{89}\)

\(\Rightarrow\)B = \(\frac{1-624}{89}=-7\)

Vậy M =\(\frac{9}{196}.-7=-\frac{9}{28}\)