Bài 1:tính
a, A = 25/6+25/6.11+25/11.16+...+25/41.46
b, B= 1/10+1/15+1/21+...+1/120
Bài 2: Chứng minh rằng:
1/2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
bài 3: tìm y
(y - 1/2).(1/2+1/6+1/12+1/20+.....+1/90) = 1/3
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18 tháng 9 2021
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
TY
11 tháng 4 2018
ta có
1/2<1/1.2
1/3<1/2.3
...
1/32<1/31.32
=>1/2+1/3+...+1/32<1/1.2+1/2.3+...+1/31.32
=>1/2+1/3+...+1/32<1/1-1/2+1/2-1/3+...+1/31-1/32
=>1/2+1/3+...+1/32<1/1-1/32=31/32
vì 31/32<1
=>tổng đó <1
ta lại có 1+1=2 mà 2 <3
=>tổng đó <3
vậy:-------(bn tự lm nha)
k cho mik vs nha
TY
11 tháng 4 2018
bn tham khảo câu trả lời của mik ở bn hoàng thu trang nhabây h mik ghi lại dài dòng lắm
NH
0
3 tháng 9 2016
a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) =
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥
♦, ♥ => 7 / 12 < A < 5 / 6
b)ta có:
1/1.2+1/3.4+1/5.6+...+1/49.50
=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)
=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2
=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)
=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50
hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
23 tháng 10 2016
\(2\)
CMR
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
b1/A=25/1.6+25/6.11+25/11.16+....+25/41.46
=5.(5/1.6+5/6.11+5/11.16+...+5/41.46)
=5.(1/1-1/6+1/6-1/11+1/11-1/16+....+1/41-1/46)
=5.(1/1-1/46)
=5.45/46
=225/46