Question

[Đặng Kiều Trang]

Chứng minh rằng:\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{49.50}\)=\(\frac{1}{26}\)+\(\frac{1}{27}\)+\(\frac{1}{28}\)+...+\(\frac{1}{50}\)

Answer 1

[Min]

\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{49.50}\)

=1-\(\frac{1}{2}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)

=(1+\(\frac{1}{3}\)+\(\frac{1}{5}\)+...+\(\frac{1}{49}\))-(\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{6}\)+...+\(\frac{1}{50}\))

=(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+...+\(\frac{1}{50}\))-2(\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{6}\)+...+\(\frac{1}{50}\))

=(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+...+\(\frac{1}{50}\))-(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+...+\(\frac{1}{25}\))

=\(\frac{1}{26}\)+\(\frac{1}{27}\)+\(\frac{1}{28}\)+...+\(\frac{1}{50}\)\(\Rightarrow\)ĐPCM

Answer 2

[Nguyen Thuy Linh]

1/1.2 + 1/3.4 + 1/5.6 +.....+1/49.50

=1- 1/2 + 1/3 - 1/4 +1/5 -1/6+....+1/49 -1/50

=(1 +1/3 +1/5 +....+1/49) - (1/2 +1/4 +1/6 +....+1/50)

=(1+1/2 +1/3 +....+1/50) - 2(1/2 + 1/4 + 1/6 +....+ 1/50)

=1+1/2 +1/3 +.....+1/50 - (1 +1/2 +1/3 +.....+1/25)

=1+1/2 +1/3 +....+1/50 -1-1/2-1/3-...-1/25

=1/26+ 1/27 +1/28 +....+1/50

Vậy 1/1.2 + 1/3.4 + 1/5.6 + .....+ 1/49.50=1/26 + 1/27 + 1/28 + ....+1/50

Mình thấy bài này dễ mà, quên mất , mình là học sinh lớp 6 đấy. Bài này như kiểu toán nâng cao lớp 6 ý. Mình nghĩ đây ko phri toán lớp 7 đâu.