QN
1
K
Khách
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XO
7 tháng 8 2020
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)
=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\)
=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)
TT
0
PT
17 tháng 4 2016
a) \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
= \(\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - \(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\) - \(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - 2.\(\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)\) - \(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
= \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\) - \(1-\frac{1}{2}-...-\frac{1}{100}\)
= \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Vậy \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\) = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Mình chỉ làm được phần a) thôi, nhưng k cho mình nhé
HC
3
LN
0
NM
1
T
2 tháng 12 2016
Xét vế phải\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}\)
=\(\left(1+\frac{1}{3}+\frac{1}{5}+..+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-...-\frac{1}{100}\)
=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
AK
28 tháng 7 2018
P/s : Đề sai mik sửa lại rồi : Tham khảo nhé :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
N
27 tháng 8 2018
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-2.\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-1+\frac{1}{2}+....+\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lời giải:
$\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{200}{101}$
$\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{n(n+1)}{2}}=\frac{200}{101}$
$\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n(n+1)}=\frac{200}{101}$
$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n(n+1)}=\frac{100}{101}$
$\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{(n+1)-n}{n(n+1)}=\frac{100}{101}$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{100}{101}$
$\frac{1}{2}-\frac{1}{n+1}=\frac{100}{101}$
$\frac{1}{n+1}=\frac{1}{2}-\frac{100}{101}=\frac{-99}{202}$
$\Rightarrow n+1=\frac{-202}{99}$ (vô lý vì $n$ là số tự nhiên.
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