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Như vậy ta sẽ so sánh 1 và 1/3 + 1/6 + 1/10 +......+ 1/45
Ta có : 1/3 + 1/6 + 1/10 + .....+ 1/45 < 1/10 + 1/10 + 1/10 +......+ 1/10
Mà 1/10 + 1/10 + 1/10 + ....+ 1/10 = 8/10 < 1
Vậy S <2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
K CHO MK NHA
Giải
S=1+1/3+1/6+1/10+...+1/45
=> S=2/2+2/6+2/12+2/20+.......+2/90
=>S=2/1x2 + 2/2x3 +2/3x4+......+2/9x10
=>S=2x(1/1x2 + 1/2x3 +....+1/9x10)
=>S=2x(1/1 - 1/2 + 1/2 - 1/3 +......+ 1/9 - 1/10
=>S=2x(1/1 - 1/10)
Vì 1/1-1/10<1=>2x(1/1 - 1/10)>2x1=2
Hay S<2
.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\frac{S}{2}=2+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}\)
\(\Rightarrow\frac{S}{2}=2+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{S}{2}=2+\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{S}{2}=2+\frac{2}{5}=\frac{12}{5}\)
\(\Rightarrow\frac{S}{2}=\frac{6}{5}< 2\)
=> S<2
\(S=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
\(S=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{90}\)
\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\)
\(S=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(S=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(S=2\left(1-\frac{1}{10}\right)\)
có : \(1-\frac{1}{10}< 1\Rightarrow2\left(1-\frac{1}{10}\right)< 2.1\)
Vậy: \(S< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(p=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+\frac{4}{46}+...+\frac{48}{2}+\frac{49}{1}\)
\(p=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+\left(\frac{4}{46}+1\right)+...+\left(\frac{48}{2}+1\right)+1\)
(do ta tách số 49 thành tổng của 49 số 1, sau đó nhóm mỗi phân số trên với 1)
\(p=\left(\frac{1}{49}+\frac{49}{49}\right)+\left(\frac{2}{48}+\frac{48}{48}\right)+\left(\frac{3}{47}+\frac{47}{47}\right)+\left(\frac{4}{46}+\frac{46}{46}\right)+...+\left(\frac{48}{2}+\frac{2}{2}\right)+1\)
\(p=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+\frac{50}{46}+...+\frac{50}{2}+1\)
\(p=50.\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+...+\frac{1}{2}\right)+1=50.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+1=50.s+1\)=> p = 50.s + 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(K=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\)
\(K=1+\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+\frac{1}{\frac{4\cdot5}{2}}+....+\frac{1}{\frac{9\cdot10}{2}}\)
\(K=1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\right)\right]\)
\(K=1+\left[2\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(K=1+\left(2\cdot\frac{2}{5}\right)=1+\frac{4}{5}=\frac{9}{5}\)
Vì \(\frac{9}{5}< 2\)\(\Rightarrow K< 2\)
\(K=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(K=1+\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(K=1+\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(K=1+\frac{2}{5}\)
\(K=\frac{7}{5}\)
S = 2/2 + 2/6 + 2/12 + 2/20 + ... + 2/90
S = 2/1.2 + 2/2.3 + 2/3.4 + .. + 2/9.10
=> S = 2( 1/1.2 + 1/2.3 + ... + 1/9.10)
=> S = 2 ( 1/1 - 1/2 + 1/2 - 1/3 + .. + 1/9 - 1/10 )
=> S = 2 ( 1/1 - 1/10 )
Vì 1/1 - 1/10 < 1 => 2 ( 1/1 - 1/0 ) < 2.1 = 2
VẬy S < 2
tick đúng nha