Chứng minh rằng 1/2.3+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

Đặt A=1/2.3+1/3.4+1/4.5+...+1/49.5, B=1/26+1/27+1/28+...+1/50

Ta có A=1/2.3+1/3.4+1/4.5+...+1/49.50

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

=1/2-1/50=25/50-1/50=24/50=12/25

Bạn ơi mình chỉ tính được cái A thôi còn cái B để mình suy nghĩ lại rồi mình giải cho

$\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\\=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\\=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\\=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\\=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\\$
Vậy `1/1.2+1/3.4+1/5.6+.......+1/49.50=1/26+1/27+....+1/50`