B = 2/3 + 2/15 + 2/35 + ......... + 2/9999 + 2/10403. C = 3/2 + 2/8 + 3/24 + ......... + 3/9800 + 3/10200. Các Bạn Jup Mình vs
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N
5 tháng 8 2017
\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{9999}+\dfrac{2}{10403}\)
\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}+\dfrac{2}{101.103}\)\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{101}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
tik mik nha !!
ND
Nguyễn Đức Trí
VIP
24 tháng 7 2023
\(T=3+8+15+24+35+...+9800+9999\) (sửa 3+5 thành 3+8)
Ta thấy :
\(3=2^2-1\)
\(8=3^2-1\)
\(15=4^2-1\)
\(24=5^2-1\)
\(.....\)
\(9800=99^2-1\)
\(9999=100^2-1\)
\(\Rightarrow T=1^2+2^2+3^2+...+100^2+\left(-1\right).100\)
\(\Rightarrow T=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-100\)
\(\Rightarrow T=\dfrac{100.101.201}{6}-100=338350-100=338250\)
5 tháng 8 2017
Ta có :
\(\dfrac{3}{2}\) = \(\dfrac{3}{1.2}\) = 3 x \(\dfrac{1}{1.2}\) = 3 x ( 1 - \(\dfrac{1}{2}\) ) : 2 ;
\(\dfrac{3}{8}\) = \(\dfrac{3}{2.4}\) = \(\)3 x \(\dfrac{1}{2.4}\) = 3 x ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ) : 2 ;
\(\dfrac{3}{24}\) = \(\dfrac{3}{4.6}\) = 3 x \(\dfrac{1}{4.6}\) = 3. ( \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) ) : 2 ;
\(\dfrac{3}{9800}\) = \(\dfrac{3}{98.100}\) = 3 x \(\dfrac{1}{98.100}\) = 3 x ( \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) ) : 2 ;
\(\dfrac{3}{10200}\) = \(\dfrac{3}{100.102}\) = 3 x \(\dfrac{1}{100.102}\)= 3 x ( \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) )
Vậy : \(\dfrac{3}{2}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{24}\) + ........ + \(\dfrac{3}{9800}\) + \(\dfrac{3}{10200}\)
= \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.4}\) + \(\dfrac{3}{4.6}\) +........+ \(\dfrac{3}{98.100}\) + \(\dfrac{3}{100.102}\)
= 3 x ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + .........+ \(\dfrac{1}{98.100}\) + \(\dfrac{1}{100.102}\) )
= 3 x ( \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) +.................+ \(\dfrac{2}{98.100}\) + \(\dfrac{2}{100.102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) +.......+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) + \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{102}\) ) : 2 = 3 x \(\dfrac{101}{102}\) : 2
= \(\dfrac{101}{68}\)
VE
19 tháng 4 2017
a, 48,63,80,99...
b, 288,399...
c,21,28,36,45,55,66,78,91...
d, 37,50,64,79,95...
X
14 tháng 2 2019
c)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)
\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có 7 số 1)
\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(7+1-\frac{1}{8}=\frac{63}{8}\)
Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé
Chúc bạn học tốt !!!
24 tháng 8 2020
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
=> \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
=> \(1-\frac{1}{2n+1}=\frac{50}{51}\)
=> \(\frac{1}{2n+1}=1-\frac{50}{51}=\frac{1}{51}\)
=> 2n + 1 = 51
=> 2n = 50
=> n = 25
Vậy n = 25
B = \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\) + ..... + \(\frac{2}{99.101}\)+ \(\frac{2}{101.103}\)
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/99 - 1/101 + 1/101 - 1/103
= 1- 1/103 = 102/103