cmr : P= 1/2^2+1/3^2+1/4^2+....+1/100^2<1
giúp mình đi các pạn ơi đắm đuối chuối luối zồi
CMR:
a)1/10^2 +1/11^2+1/12^2+...+1/100^2 >3/4
b)1/2^2+1/3^2+1/4^2+...+1/100^2<99/100
c)1/2^2+1/3^2+1/4^2+...+1/100^2<3/4
CMR:(1+1/2+1/3+1/4+...+1/100)=1/2=2/3+3/4+...+99/100
cmr
100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+....+99/100
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
CMR
1-1/2^2-1/3^2-1/4^2-....-1/100^2<1/100
CMR
100-(1/2+1/3+1/4+...+1/100)=2/3+3/4+...+99/100
100=10*10
100=1000:10
100 câu nói hay về cuộc sống
cmr :1/2^2+1/4^2+...+1/100^2<1
1/2!+1/3!+1/4!+...+1/100!<1
B=1/2^2+1/3^2+1/4^2+...+1/100^2 CMR B<3/4
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\frac{1}{1\cdot2}< \frac{1}{2^2}\)
\(\frac{1}{2\cdot3}< \frac{1}{3^2}\)
\(\frac{1}{3\cdot4}< \frac{1}{4^2}\)
...
\(\frac{1}{99\cdot100}< \frac{1}{100^2}\)
\(\Rightarrow B< A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(B< A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}>\frac{3}{4}\)
\(\Leftrightarrow B< \frac{3}{4}< A\)
CMR
A=1/2^2+1/3^2+1/4^2+......+1/100^2 <3/4
TA CÓ 1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
1/100^2<1/99.100
=>1/2^2+2/3^2+.....+1/100^2<1/1.2+1/2.3+..+1/99.100
=1-99/100=99/100<1
CMR : a) 1/2! + 2/3! + 3/4! +...+ 99/100! < 1
b) 1.2-1/2! + 2.3-1/3! + 3.4-1/4! +...+ 99.100-1/100! < 2
\("!"\) là giai thừa đó bạn ạ .
\(VD:\) \(3!=1.2.3=6\)
\(4!=1.2.3.4=24\)
cmr P= 1/2^2 + 1/3^2 +1/4^4 + .... + 1/100^2