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P2
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
12 tháng 12 2019
Nếu \(n>0\Rightarrow\left(n-1\right)n\left(n+1\right)=n^3-n< n^3.\)
\(\Rightarrow VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2007-2005}{2005.2006.2007}\)
\(\Rightarrow2VT< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(\Rightarrow2.VT< \frac{1}{2}-\frac{1}{2006.2007}\Rightarrow VT< \frac{1}{4}-\frac{1}{2.2006.2007}< \frac{1}{4}\)
HT
1
7 tháng 1 2016
XET 1+1/2^2+1/3^2+1/4^2+....+1/100^2
=>1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1 + 1/1.2 + 1/2.3 +... + 1/100.101
=> 1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1+ (1-1/2+1/2-1/3+...+1/100-1/101)
=> 1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1+ 1-1/101 <2
TN
0
PM
2
8 tháng 10 2018
Ta có : \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B=\frac{2015}{51}+\frac{2015}{52}+...+\frac{2015}{100}\)
\(=2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow\) \(\frac{B}{A}=\frac{2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2015\)
\(\Rightarrow\) \(B⋮A\)
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
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)