CM
1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2 < 1
Những câu hỏi liên quan
cho x+y+z=o,xyz=0
cm1/x^2+y^2-z^2+1/y^2+z^2-x^2+1/z^2+x^2-y^2=0
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
1. (1+1/2).(1+1/2^2).(1+1/2^3)....(1+1/2^100) < 3
2. 1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+ 1024/(5^1024+1) <1/4
3. 3/(1!+2!+3!)+4/(2!+3!+4!)+...+100/(98!+99!+100!) <1/2
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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!
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1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow A< 1\)
Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)
\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)
*Cách khác:
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Ta thấy:
\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)
\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Mà \(\frac{2+1}{2}< 3\)
\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)
Xem thêm câu trả lời
Tính:
M=(1-1/2^2).(1-1/3^2).(1-1/4^2)...(1-1/49^2).(1-1/50^2)
N=(3/2-2/2^2).(4/3-2/3^2).(5/4-2/4^2)...(100/99-2/99^2).(101/100-2/100^2)
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bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
Tính:
A=(1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4)...(1-1/1+2+3+4+...+2022)
B=1+1/2(1+2)+1/3(1+2+3)+1/100(1+2+3+...+100)
Tính C=1/2-(1/3+2/3)+(1/4+2/4+3/4)-(1/5+2/5+3/5+4/5)+...+(1/100+2/100+...+99/100)
Tính tổng 100-(1+1/2+1/3+1/4+...+1/100)/1/2+2/3+3/4+....+99/100
A = \(\dfrac{100-(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{99}{100}}\)
Xét các mẫu số của dãy phân số : \(\dfrac{1}{1};\dfrac{1}{2};....;\dfrac{1}{100}\)
ta có dãy số: 1; 2; ....;100
Dãy số trên có số số hạng là: ( 100 - 1) : 1 + 1 = 100 (số)
Tách 100 thành tổng của 100 số 1 rồi nhóm lần lượt 1 với từng phân số thuộc dãy phân số trên khi đó ta có:
A = \(\dfrac{100-(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.....+\dfrac{99}{100}}\)
A = \(\dfrac{(1-1)+(1-\dfrac{1}{2})+(1-\dfrac{1}{3})+....+(1-\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.....+\dfrac{99}{100}}\)
A = \(\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)
A = 1
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(1/100-1/2^2).(1/100-1/3^2).(1/100-1/4^2)........(1/100-1/2022^2)
M=1 + 1/2 (1+2) + 1/3 (1+2+3) +1/4 (1+2+3+4) +...+ 1/100. (1+2+3+...+100) = ?