S=1/2²+1/3²+1/4²+1/5²+...+1/100²<1 help me với ạ
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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, S= 1/1*2 + 1/2*3 + 1/3*4 +...+1/99*100
S= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100
S= 1/1 - 1/100
S= 100/100 - 1/100
S= 99/100
b, S= 1/1*3 + 1/3*5 + 1/5*7 +...+1/99*101
S= 1/2* (2/1*3 + 2/3*5 + 2/5*7 +...+ 2/99*101)
S= 1/2* (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +...+ 1/99 - 1/101)
S= 1/2* (1/1 - 1/101)
S= 1/2* (101/101 - 1/101)
S= 1/2* 100/101
S= 50/101
Chúc bạn học tốt nha
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bạn ấn vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm
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S1=1+2+3+...+999
Số số hạng S1= (999-1):1+1=999(số hạng)
tổng S1= \(\left(999+1\right)+\left(998+2\right)+...+\left(499+501\right)+500\)
\(=\left(999+1\right).499+500\)
\(=499500\)
S2=1-2+3-4+...+99-100+101
=(1-2)+(3-4)+...+(99-100)+101
=(-1)+(-1)+...+(-1)+101
=(-1).50+101
=(-50)+101
=51
![](https://rs.olm.vn/images/avt/0.png?1311)
S= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=1-\frac{1}{100}\)
\(S=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
`5S=5(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`5S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)`
`=>5S-S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)-(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`4S=1/5+1/(5^2)+1/(5^3)+1/(5^4)+...+1/(5^99) -99/(5^100)`
`20S=5(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`20S=1+1/5+1/(5^2)+....+1/(5^98)-99/(5^99)`
`=>20S-4S=(1+1/5+1/(5^2)+...+1/(5^98)-99/(5^99))-(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`=>16S=1-99/(5^99)-1/(5^99)-99/(5^100)`
Vì `-99/(5^99)-1/(5^99)-99/(5^100)<0=>1-99/(5^99)-1/(5^99)-99/(5^100)<1`
`=>S<1/16`
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(S< 1-\dfrac{1}{100}< 1\)