cho A= 1/3+1/6+1/10+....+1/4950.So sánh A với 1/4
K
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 3 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + .....+ \(\dfrac{1}{4950}\)
A = \(\dfrac{2}{2}\) \(\times\) ( 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+.......+ \(\dfrac{1}{4950}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+......+ \(\dfrac{1}{9900}\))
A = 2 \(\times\) ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+....+ \(\dfrac{1}{99.100}\))
A = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
A = 2 \(\times\) ( 1 - \(\dfrac{1}{100}\))
A = 2 \(\times\) \(\dfrac{99}{100}\)
A = \(\dfrac{99}{50}\)
NC
28 tháng 8 2020
Bài làm:
Ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left[\left(1+\frac{1}{3}+...+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]-\left[\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)=B\)
Vậy A = B
TT
1
MA
30 tháng 3 2016
nhân 2 vế với 1/2 ta có
1/2 x A = 1/2 x (1/3 + 1/6 +1/10 + 1/15 + .......+1/91 + 1/105 )
1/2 x A = 1/6 +1/12 + 1/20 + 1/30 + ...............+1/182 + 1/210
1/2 x A = 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) +................+1/(13x14) + 1/(14x15)
1/2 x A = 1/2 - 1/3 +1/3 -1/4 + 1/4 - 1/5 +1/5 - 1/6+.........+1/13 - 1/14 + 1/14 - 1/15
1/2 x A = 1/2 - 1/15 =13/30
=> A = 13/30 : 1/2=13/15 <1
PN
10 tháng 5 2021
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
DL
1
TA
18 tháng 3 2018
tổng của A là
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 223/140
=> 223/140 > 6/7
k mk nha
1 tháng 5 2017
Ta có :
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+\left(\frac{4}{4!}-\frac{1}{4!}\right)+...+\left(\frac{10}{10!}-\frac{1}{10!}\right)\)
\(A=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{9!}-\frac{1}{10!}\right)\)
\(A=1-\frac{1}{10!}< 1\)
vậy A < 1 vì \(0< \frac{1}{10!}< 1\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)