Tính S=1+4.5+7.52+....+288.599
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\(S=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.......+\frac{1}{49\cdot50}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{49}+\frac{1}{50}\)
\(S=\frac{1}{2}-\frac{1}{50}\)
\(S=\frac{25}{50}-\frac{1}{50}\)
\(S=\frac{24}{50}=\frac{12}{25}\)
ai k mh mh k lại
k cho mh nha
S=1/2.3+1/3.4+1/4.5+....+1/49.50
=\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{49x50}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{2}-\frac{1}{50}\)
=\(\frac{24}{50}\) mình cũng ko chắc đúng nhưng đây là cách giải của mình
\(S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{2}-\frac{1}{100}\)
\(S=\frac{49}{100}\)
chúc các bạn học tốt
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=1\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=1\times\frac{49}{100}\)
\(S=\frac{49}{100}\)
s = 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5
S=1 + (-1/2 +1/2)+...+(-1/4 + 1/4 ) +-1/5
S = 1 + 0 +0 +...+ 0 +-1/5
S= 1 + -1/5
S = 4/5
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
S=1.2+2.3+3.4+...+99.100
3S=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3S=99.100.101
S=(99.100.101):3=333300
S = 1/2 - 1/3 + 1/3 -1/4 + ......... +1/2011 -1/2012
S= 1/2 - 1/2012 = 1005/2012
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{2012}\)
\(S=\frac{1}{2}+0+0+0+...-\frac{1}{2012}\)
\(S=\frac{1}{2}-\frac{1}{2012}\)
\(S=\frac{1005}{2012}\)
\(A=\frac{2012}{1}\cdot\frac{1005}{2012}\)
\(A=1005\)
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101
=> 3S = 99.100.101
=> S = \(\frac{99.100.101}{3}=333300\)
ta xét
\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)
\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)
\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)
Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)
`S = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100.`
`3S = 1.2.3 + 2.3.(4-1) + 3.4.(5-4) + 4.5.(6-3) + ... + 99.100.(101-98)`
`3S = 1.2.3 + 2.3.4-1.2.3 + 3.4.5-4.5.6 + 4.5.6-3.4.5 + ... + 99.100.101-98.99.100`
`3S = 99.100.101`
`S = 33.100.101`
`S = 333300`
3S=1.2(3-0)+2.3(4-1)+.....+99.100(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-2.3.4+....+99.100.101-98-99-100
=99.100.101
S=33.100.101
=333300