Tính Tổng
A=1+6+11+16+21+.....+101
B=1.2+2.3+3.4+....+98.99
C=\(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....+\frac{1}{3^{100}}\)
K
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11 tháng 8 2016
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)
=> \(\frac{1}{x}=3\)
=> \(x=\frac{1}{3}\)
26 tháng 6 2019
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
23 tháng 8 2019
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
23 tháng 8 2019
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
18 tháng 2 2017
a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)
b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)
\(=0+\frac{-125}{143}=-\frac{125}{143}\)
18 tháng 2 2017
bài 2
a \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(1-\frac{1}{2004}=\frac{2003}{2004}\)
NH
1
HT
8 tháng 9 2015
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
= \(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+....+\frac{99.100}{100!}-\frac{1}{100!}\)
= \(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
= \(\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}\right)\)
= \(1+1-\frac{1}{99!}\)
= \(2-\frac{1}{99!}
11 tháng 9 2016
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\left(đpcm\right)\)
27 tháng 8 2017
a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
Q
27 tháng 8 2017
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
9 tháng 6 2017
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
PG
0
*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071