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\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
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
Có \(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{43.46}\)
Sẽ có: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{43.46}\)
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+........+\frac{1}{43}-\frac{1}{46}\)
\(S=\frac{1}{1}-\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+....+\left(-\frac{1}{43}+\frac{1}{43}\right)-\frac{1}{46}\)
\(S=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{46}\)
\(S=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)
Vì có: \(\frac{45}{46}<1\) nên \(S<1\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
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
S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46
S= 1-1/46
S= 45/46<1
vậy S<1
duyệt đi
S= 1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46
S= 1-1/46= 45/46<1
Suy ra S<1
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
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
Bài làm :
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{43-40}{40.43}+\frac{46-43}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}< 1\)
=> Điều phải chứng minh
S=1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1-1/46 S=45/46<1
vậy S <1
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)
Vậy S < 1 (đpcm)
