Chứng minh rằng :
$\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ + … + $\frac{1}{49.50}$ = $\frac{1}{26}$ + $\frac{1}{27}$ + $\frac{1}{28}$ + … + $\frac{1}{50}$
Chứng minh rằng :
$\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ + … + $\frac{1}{49.50}$ = $\frac{1}{26}$ + $\frac{1}{27}$ + $\frac{1}{28}$ + … + $\frac{1}{50}$
Đáp án: $\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+…+\dfrac{1}{50}$
Giải thích các bước giải:
Ta có: $\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{49.50}$
$=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+…+\dfrac{1}{49}-\dfrac{1}{50}$
$=\left ( 1+\dfrac{1}{3}+\dfrac{1}{5}+…+\dfrac{1}{49} \right )-\left ( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+…+\dfrac{1}{50} \right )$
$=\left ( 1+\dfrac{1}{3}+\dfrac{1}{5}+…+\dfrac{1}{49} \right )+\left ( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+…+\dfrac{1}{50} \right )-2\left ( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+…+\dfrac{1}{50} \right )$
$=\left ( 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+…+\dfrac{1}{49}+\dfrac{1}{50} \right )-\left ( 1+\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{25} \right )$
$=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+…+\dfrac{1}{50}$
Vậy $\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+…+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+…+\dfrac{1}{50}$
Đáp án:
1/1.2+1/3.4+1/5.6+…+1/49.50=1/26+1/27+1/28+…+1/50
= 1/1-1/2+1/3-1/4+…+1/50
= (1/1+1/3+…+1/49)-(1/2+1/4+…+1/50)
= (1/1+1/2+1/3+…+1/49+1/50)-2(1/2+1/4+…+1/50)
=1/1+1/2+1/3+…+1/49+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
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