1/1×2+1/2×3+1/3×4+1/4×5+...+1/49×50
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NP
0
26 tháng 9 2015
1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ... +1/49*50
= 1 - 1/2 + 1/2- 1/3 + ... + 1/49 - 1/50
= 1 - 1/50
= 49/50
26 tháng 9 2015
1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ... +1/49*50
1-1/2 +1/2 -1/3+ .... +1/49 -1/50
=1-1/50
=49/50
15 tháng 4 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{50}\)
\(\frac{49}{50}\)
KN
15 tháng 4 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
VP
0
26 tháng 1 2023
So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2
NC
26 tháng 1 2023
S=
=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50
P=
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50=1
vậy s/p = 1/50
VP
2
29 tháng 7 2015
Ax2=1x2/1x2x3+1x2/2x3x4+...+1x2/48x49x50
Ax2=1/1x2-1/2x3+1/2x3-1/3x4+...+1/48x49-1/49x50
Ax2=1/1x2-1/49x50
Ax2=1/2-1/2450
Ax2=1225/2450-1/2450
Ax2=1224/2450
A=1224/2450:2
A=1224/2450X1/2
A=1224/4900
A=306/1225
NT
0
KT
3
25 tháng 8 2021
Xét vế phải :
\(VT=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)