Chứng minh A-38B=0 biết A=1/1.2+1/3.4+...+1/49.50 và B=1/26.50+1/27.49+...+1/50.51

Dạng hay :v

Ta có:
\(A = \dfrac{1}{1.2} + \dfrac{1}{3.4} +...+ \dfrac{1}{49.50}\)
\(=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{49})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50})\)
\(=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50})-2.(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50})\)
\(=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50})-(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25})\)
\(=>A=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50} (1)\)
Ta lại có:
\(B = \dfrac{1}{26.50} + \dfrac{1}{27.49} +...+ \dfrac{1}{50.26}\)
\(=>38B=\dfrac{38}{26.50}+\dfrac{38}{27.49}+...+\dfrac{38}{50.26}\)
\(=>38B=\dfrac{76}{26.50}+\dfrac{76}{27.49}+...+\dfrac{38}{38.38}\)
\(=>38B=\dfrac{1}{26}+\dfrac{1}{50}+\dfrac{1}{27}+\dfrac{1}{49}+...+\dfrac{1}{38}\)
\(=>38B=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50} (2)\)
Từ (1)(2):
\(=>A = 38B\)
\(=>A-38B=0\)

Ta có: 
$A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}$
$=>A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}$
$=>A=(1+\frac{1}{3}+...+\frac{1}{49})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50})$
$=>A=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50})-2.(\frac{1}{2}+\frac{1}{4}+.$

  bởi Hacker mũ trắng 15/04/2019

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