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Ta có: A=1.2+2.3+...+198.199+199.200
=>3A=1.2.3+2.3.3+...+198.199.3
+199.200.3
=>3A=1.2.3+2.3(4-1)+...+
198.199(200-197)+199.200(201-198)
=>3A=1.2.3+2.3.4-1.2.3+...+198.199.200
-197.198.199+199.200.201-198.199.200
=>3A=199.200.201
=>A=199.200.67
A=39800.67
A=2666600
...
= 1/2-1/3+1/3-1/4+...+ 1/19-1/20
= 1/2-1/20
=9/20
có phải như thế này ko bn
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)
A = \(\frac{9}{20}\)
\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)
B = \(-\frac{99}{100}\)
có: 1/1.2= 1/1-1/2
1/2.3=1/2-1/3
...
1/8.9= 1/8-1/9
1/9.10=1/9-1/10
suy ra B=1/1-1/2+1/2-1/3+...+1/9-1/10
=1/1-1/10 (vì bạn thấy trừ 1/2 rồi cộng 1/2, trừ 1/3 cộng 1/3,... thì khử đi)
=9/10.
vậy B=9/10
1.50+2.49+3.48+...+49.2+50.1=
= (1.50+2.50+3.50+...+50.1)-(1.2+2.3+3.4+...+49.50)
= (2500+50).50:2-41650
= 63750-41650=22100
2,
A = 1.2 + 2.3 + 3.4 + ... + 2011.2012
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012
3A = 2011.2012.2013
A = 2011.2012.2013 : 3
A = 2714954572
\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\
=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\
=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\
=-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)
Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
=>\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=>\(A=1-\frac{1}{50}\)
=>\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}\)
3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3
2013/2014
\(\frac{2013}{2014}\)