CMR
1/26 + 1/27 + 1/28+....+ 1/50 = 1 - 1/2 + 1/3 - 1/4 + ... +1/49 - 1/50
NN
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CMR: 1/26 + 1/27 + 1/28 + ... + 1/50 = 1 - 1/2 + 1/3 - 1/4 + ... + 1/49 - 1/50
gấpp ạaaa
Lời giải:
Ta có:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}$
$=(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{49}+\frac{1}{50})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{49}+\frac{1}{50})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25})$
$=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}$
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Tính : 1/26+1/27+1/28+.......+1/50=1-1/2+1/3-1/4+.....+1/491/26+1/27+1/28+.......+1/50=1-1/2+1/3-1/4+.....+1/49
CMR
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{60}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)
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CMR:\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
H24
28 tháng 6 2021 lúc 14:24
Chứng minh rằng 1/26 + 1/27 + 1/28 +...+ 1/50 = 1 - 1/2 + 1/3 - 1/4 +...+ 1/49 - 1/50
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)(đpcm)
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\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\) (đpcm)
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Giải:
\(\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
Ta có:
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\left(đpcm\right)\)
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CM: 1/26+ 1/27+ 1/28+....+ 1/50= 1- 1/2+ 1/3- 1/4+...+ 1/49- 1/50
CMR:
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\)
C/m : 1/26+1/27+1/28+...+1/50=1-1/2+1/3-1/4+...+1/49-1/50
1/1*2+1/3*4+1/5*6+...+1/49*50=1/26+1/27+1/28+...+1/50