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![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+.........+\left(1+2+3+......+98\right)}{1.2+2.3+3.4+.............+98.99}\) \(A=\dfrac{1+3+6+................+4851}{2+6+12+..........+9702}\)
\(A=\dfrac{1+3+6+..........+4851}{1.2+2.3+2.6+........+2.4851}\)
\(A=\dfrac{1}{2}\)
Vậy\(A=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta co \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^{10}}\)
=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
=\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
=\(\frac{1}{1^2}-\frac{1}{10^2}\)
=\(\frac{99}{100}\) < 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Đặt \(A=1.2+2.3+3.4+....+98.99\)
Ta có:\(3A=3.\left(1.2+2.3+3.4+....+98.99\right)\)
\(3A=1.2.3+2.3.3+3.4.3+....+98.99.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}=323400\)
Ta có:\(\frac{A.y}{1}=184800\Rightarrow y=184800:323400=\frac{4}{7}\)
2)Đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185,8\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\)
Tổng quát:\(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right)a}-\frac{1}{a\left(a+1\right)}\)
Ta có:
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{37.38.39}\)
\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\Rightarrow B=\frac{370}{741}:2=\frac{185}{741}\)
Khi đó \(A=\frac{185}{741}.1428+185,8=...........\) (tự tính ra)
(*)số ko đẹp mấy
![](https://rs.olm.vn/images/avt/0.png?1311)
B= 333300
C=328350
D=(n+1) /( n nhân 2)
E=(1/3 trừ 1/3^100):2
1)=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3
3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3B=99.100.101
=>B=333300
\(2A=\frac{1.2+2.3+3.4+...+98.99}{1.2+2.3+3.4+...+98.99}\)
\(2A=1\)
\(A=\frac{1}{2}\)