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S=\(\frac{1}{1}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{10}\)+...+\(\frac{1}{100}\)-\(\frac{1}{103}\)+\(\frac{1}{103}\)-\(\frac{1}{104}\)+\(\frac{1}{104}\)-\(\frac{1}{105}\)+\(\frac{1}{105}\)-\(\frac{1}{106}\)+\(\frac{1}{106}\)-\(\frac{1}{107}\)
S=1-\(\frac{1}{107}\)
S=\(\frac{106}{107}\)
(Ở đề bài, ở phân số cuối cùng 1/106+107 nên sửa lại thành 1/106.107 sẽ được kết quả như trên)
Ta có: \(S=\frac{1}{1}-\frac{1}{103}+\frac{1}{103}-\frac{1}{107}\)
\(S=1-\frac{1}{107}=\frac{106}{107}\)
Trong trường hợp bn viết nhầm 1/106.107 chứ ko phải 1/106+107
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106.107}\)
\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}+\frac{1}{106}-\frac{1}{107}\right)\)
\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{107}\right)\)
\(S=\frac{102}{103}+\frac{4}{11021}\)
\(S=\frac{106}{107}\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}+\frac{1}{103.104}+\frac{1}{104.105}+\frac{1}{105.106}+\frac{1}{106+107}\)
\(S=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}+\frac{1}{105}-\frac{1}{106}\right)+\frac{1}{106+107}\)
\(S=\left(1-\frac{1}{103}\right)+\left(\frac{1}{103}-\frac{1}{106}\right)+\frac{1}{106+107}\)
\(S=\frac{102}{103}+\frac{3}{10918}+\frac{11343}{106}\)
\(S=108\)
\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\frac{102}{103}\)
\(B=\frac{34}{103}\)
Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)
\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)
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= 1-1/4+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
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=(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)+(1/103-1/104+1/104-1/105+1/105-1/106+1/106-1/107)
=(1-1/103)+(1/103-1/107)
=1 - 1/107
=106/107