Tính nhanh:
1/26+1/27+1/28+...+1/50
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Ta có:
\(M=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(M=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=N\)
\(\Rightarrow\frac{M}{N}=1\)
Ta biến đổi vế phải :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\\ \)\(\\ =\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}\)
Vậy \(\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
=> \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\) ( đpcm )
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)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\) (đpcm)
Lời giải:
Ta có:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}$
$=(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{49}+\frac{1}{50})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{49}+\frac{1}{50})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25})$
$=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}$
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(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+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
1/26+1/27+1/28+...+1/49+1/50
2/26+2/28+2/30+...+2/50
<=>1/13+1/14+1/15+...+1/25
<=>2/14+2/16+2/18+...2/24
<=>1/7+1/8+1/9+...+1/12
<=>2/8+2/10+2/12
<=>1/4+1/5+1/6
<=>2/4+2/6
<=>1/2+1/3
<=> 2/2 = 1
1/26+1/27+1/28+...+1/49+1/50
=1-1/2+1/3-1...2/26+2/28+2/30+...+2/50=1-1/2+1/3-1...
=1/13+1/14+1/15+...+1/25=1-1/2+1/3-1...
=2/14+2/16+2/18+...2/24=1-1/2+1/3-1/...
=1/7+1/8+1/9+...+1/12=1-1/2+1/3-1/4+...
=2/8+2/10+2/12=1-1/2+1/3-1/4+1/5-1/6
=1/4+1/5+1/6=1-1/2+1/3-1/4+1/5-1/6
=2/4+2/6=1-1/2+1/3
=1/2+1/3=1-1/2+1/3
= 2/2 = 1