ChoN=1/2+(1/2)^2+(1/2)^3+(1/2)^4+......+(1/2)^98+(1/2)^99. Chứng minh B<1
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Chứng minh 1/2+(1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^98+(1/2)^99 < 1
A=(1/2+1/2)^2+(1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99
Chứng minh A<1
1. Cho B= (1/2) + (1/2)2 + (1/3)3 + (1/2)4 + ... + (1/2)98 + (1/2)99
Chứng minh: B<1
B=1/2 +(1/2 )^2+(1/3 )^3+......+(1/2 )\(^{99}\)
⇒2B=1+1/2 +1/22 +......+1/298
⇒B=2B−B=1−1/2\(^{99}\)
⇒1−1/2\(^{99}\) <1⇒B<1
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\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=> \(2B-B=\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{98}\right)\)\(-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
=> \(B=1-\frac{1}{2^{99}}< 1\)
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\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+......+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)
\(\Rightarrow B=2B-B=1-\frac{1}{2^{99}}\)
\(\Rightarrow1-\frac{1}{2^{99}}< 1\Rightarrow B< 1\)
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1. Tìm 2 số nguyên tố x và y sao cho
x2 -2x+1=6y2-2x+2
2. a/b=1/50+1/51.....1/99 CHỨNG MINH a chia hết cho 149
3. Cho m=(1/1+1/2+1/3....+1/98)*2*3.....*98 CHỨNG MINH m chia hết cho 99
a) thu gọn biểu thức sau: a= 5 - 5^2 + 5^3 - 5^4 +...- 5^98 + %^99
b) chứng minh rằng với mọi n thuộc N thì (2^n+1).(2^n+2) đều chia hết cho 3
c) chúng minh: A= 1/1^2 + 1/2^2+ 1/3^2+.....+1/99^2+ 1/100^2 < 1 3/4 (hỗn số)
cho B=1/2+(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99
chứng minh rằng B<1
\(\frac{B}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\frac{B}{2}=B-\frac{B}{2}=\frac{1}{2}-\frac{1}{2^{100}}< 1\)
Cho A=[1/1+1/2+1/3+...+1/98]*2*3*4*...*98
Chứng minh A chia hết cho 99
A=[1/1+1/2+....+1/98]*2*4*...*98*3*33=A=[1/1+1/2+....+1/98]*2*4*....*98*99\(⋮\)99
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\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times3\times4\times...\times98\)
\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times3\times4\times...\times33\times...\times98\)
\(A=\left(3\times33\right)\times\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times4\times...\times98\)
\(A=99\times\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times4\times...\times98\)
Vậy \(A⋮99\)(Vì A có thừa số 99)
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B= 1/3 + 1/3^2 +...+ 1/3^98 + 1/3^99
Chứng minh B<1/2
B = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow\)3B = \(1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
Lấy 3B - B = \(\left(1+\frac{1}{3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
2B = \(1-\frac{1}{3^{99}}\)
B = \(\left(1-\frac{1}{3^{99}}\right):2\)
= \(\left(1-\frac{1}{3^{99}}\right).\frac{1}{2}\)
= \(1.\frac{1}{2}-\frac{1}{3^{99}}.\frac{1}{2}\)
= \(\frac{1}{2}-\frac{1}{3^{99}.2}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\left(đpcm\right)\)
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Cho A = ( 1 + 1/2 + 1/3 + ... + 1/98 ) x 2 x 3 x 4 x ... x 98
Chứng minh A chia hết cho 99