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\(VT=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+\frac{1}{3!}-\frac{1}{5!}+...+\frac{1}{97!}-\frac{1}{99!}+\frac{1}{98!}-\frac{1}{100!}\)
\(VT=2-\frac{1}{100!}< 2\)đpcm
Ta xét vế trái nha
\(VT=\frac{1.2-1}{2}+\frac{2.3-1}{3}+\frac{3.4-1}{4}+.....+\frac{99.100-1}{100}\)
\(=1-\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}......+\frac{1}{98}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=>VT< VP\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)
1) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
2)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{99^2}< \frac{1}{98.99}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(1-\frac{1}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< \frac{99}{100}< 1\)
\(\frac{98}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2.3}< \frac{1}{2^2}\)
\(\frac{1}{3.4}< \frac{1}{3^2}\)
\(...\)
\(\frac{1}{99.100}< \frac{1}{99^2}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{49}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< 1\)
a) A=(100-1):1+1=100 số hạng
A=100:2=50 cặp
tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101
tính giá trị của biểu thức A: 50*101=5050
[ mình tính theo công thức đó ]
2A=1+1/2+1/2^2+1/2^3+...+1/2^99
-A= 1/2+1/2^2+1/2^3+...+1/2^99+1/2^100
-------------------------------------------------------------------
A=1-1/2^100
A=2^100-1/2^100<1(dpcm)
B), B=2/1.2 +22.3 +23.4 +...+299.100 <2 =
=1-1/2-1/2-1/3+.........+1/99-1/100
=1-1/100
=99/100
vì 99/100<2 nên B=2/1.2+2/2.3+2/3.4+......+2/99.100<2
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}\)\(+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)\)\(-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}-...-\frac{1}{100!}\)
\(=1+1+\frac{1}{2!}+...+\frac{1}{98!}-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}-...-\frac{1}{100!}\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(=1-\frac{1}{2!}+\frac{1}{1!}-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{99!}-\frac{1}{100!}\)