Cho C = 1/100-1/100*99-....1/1*2
Khi đó 50c =
Cho C=(1/100)-(1/(100x99)) -(1/(99x98)) -...-(1/(2x1)) . Khi đó 50C=
Tính tổng sau:
a) 1 - 2 + 3 - 4 + ... + 99 - 100
b) 2 - 4 + 6 - 8 + ... + 48 - 50
c) 1 + 2 - 3 + 4 - ... - 99 + 100
a: từ 1 đến 100 sẽ có \(\dfrac{100-1}{1}+1=100-1+1=100\left(số\right)\)
=>Sẽ có \(\dfrac{100}{2}=50\) cặp số
1-2+3-4+...+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+...+(-1)
=-1*50=-50
b: Sửa đề: \(2-4+6-8+...+46-48+50\)
Từ 2 đến 48 sẽ có \(\dfrac{48-2}{2}+1=24-1+1=24\left(số\right)\)
=>Sẽ có \(\dfrac{24}{2}=12\left(cặp\right)\)
\(2-4+6-8+...+46-48+50\)
\(=\left(2-4\right)+\left(6-8\right)+...+\left(46-48\right)+50\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+50\)
\(=50-2\cdot24=50-48=2\)
c: Đặt A=\(1+2-3+4+...+97+98-99+100\)
\(=\left(1+2-3+4\right)+\left(5+6-7+8\right)+...+\left(97+98-99+100\right)\)
\(=4+12+...+196\)
Từ 4 đến 196 sẽ có \(\dfrac{196-4}{8}+1=\dfrac{192}{8}+1=25\left(số\right)\)
Tổng của dãy A là: \(\left(196+4\right)\cdot\dfrac{25}{2}=\dfrac{25}{2}\cdot200=100\cdot25=2500\)
a.1-2+3-4+......+99-100
b.2-4+6-8+......-48+50
c.1+2-3-4+.......+97+98-99-100
a: 1-2+3-4+...+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+...+(-1)
=-1*50=-50
c: 1+2-3-4+....+97+98-99-100
=(1+2-3-4)+(5+6-7-8)+...+(97+98-99-100)
=(-4)+(-4)+...+(-4)
=(-4)*25=-100
cho C = 1/100 - 1/100x99 - 1/99x98 - ........... - 1/3x2 - 1/2x1
Khi đó 50C = ?
CHỈ CẦN ĐÁP ÁN THUI NHA
C=\((^1/100)-(^1/100.99)-(^1/99.98)-(^1/98.97)-...-(^1/3.2)-(^1/2.1) \)
Khi đó : 50C = ?
Cho \(c=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\) .
Khi đó 50C =
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
=> 50C = \(50.\left(-\frac{49}{50}\right)=-49\)
Cho C =\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
Khi đó 50C =
\(C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\cdot\cdot\cdot+\frac{1}{2\cdot1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{100}{100}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=\frac{-98}{100}\)
\(Khi\) đó \(50C=\frac{-98}{100}\cdot50\)
\(50C=-49\)
tick cho mình nha
Cho C=1/100-1/100.99-1/99.98-1/98.97-...-1/3.2-1/2.1
Khi do 50C=...
Cho \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
Khi đó \(50C=\)
C = \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
\(\Rightarrow50C=50.\left(-\frac{49}{50}\right)=-49\)
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Đặt A= \(-\frac{1}{100.99}-\frac{1}{99.98}-....-\frac{1}{2.1}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
C = \(\frac{1}{100}+\left(-\frac{99}{100}\right)=-\frac{49}{50}\)
50C = \(-\frac{49}{50}.50=-49\)