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\(^{\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{10}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(1-\frac{1}{46}=\frac{45}{46}\)
Vì \(1-\frac{1}{46}< 1\)nên \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}< 1\)
Chúc bạn học tốt
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}\)
\(S=\frac{42}{43}< 1\)
Ta có :
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+..............+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...............+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}+\dfrac{1}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}=\dfrac{45}{46}\)
\(\dfrac{45}{46}< 1\)
=> \(S< 1\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
ta có
\(\frac{3}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{1}{4}-\frac{1}{7}\)
.....
\(\frac{3}{43.46}=\frac{1}{43}-\frac{1}{46}\)
( mình nghĩ cậu chưa đc làm dạng như này nên ghi ra , lần sau có gặp mà biết cách làm r thì bỏ bước trên đi cx đc nha)
\(=>S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=>S=1-\frac{1}{46}< 1\)(dpcm
Ta có\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}\)
Vì \(1-\frac{1}{46}< 1\)nên S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+.......+\frac{3}{43\cdot46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Ta có \(1-\frac{1}{46}< 1\)=> S < 1
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Có \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\)
nhan xet:3/1.4=1/1-1/4
3/4.7=1/4-1/7
3/7.10=1/7-1/10
.....................
3/40.43=1/40-1/43
3/43.46=1/43-1/46
S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=46/46-1/46
S=45/46<1
vay s<1
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Đến đây ta suy được ra S<1
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy \(S< 1\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{n\left(n+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}\)
Ta có :
\(\frac{1}{n+3}>0\)
\(\Leftrightarrow-\frac{1}{n+3}< 0\)
\(\Leftrightarrow1-\frac{1}{n+3}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+2}{n+3}\)
Vi \(n\inℕ^∗\)nên \(n+2< n+3\)
DO đó\(\frac{n+2}{n+3}< 1\)
Vậy S <1
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(s=1-\frac{1}{46}< 1\)
Vậy S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{43\cdot46}\)
\(S=1\left[\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{43\cdot46}\right]\)
\(S=1\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\right]\)
\(S=1\left[1-\frac{1}{46}\right]=1\cdot\frac{45}{46}=\frac{45}{46}< 1(đpcm)\)