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29 tháng 6 2016
\(a.A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\)
\(b.B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
Ủng hộ mk nha ^_^
28 tháng 2 2020
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
28 tháng 2 2020
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
VT
0
NM
0
25 tháng 5 2015
Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
CG
1
22 tháng 8 2015
p = (1 + 31) + (32 + 33) + ...+ (398 + 399) = 4.1 + 32.(1+ 31) + ...+ 398.(1+ 31) = 4.1 + 32.4 + ....+ 398.4
= 4. (1 + 32 + 34 + ...+ 398) chia hết cho 4
=> p chia hết cho 4
p = (1+ 31+ 32 + 33) + (34 + 35 + ...+ 37) + ...+ (396 + 397 + 398 + 399)
= 40 + 34.(1 + 31 + 32+ 33) + ....+ 396. (1+31 + 32 + 33)
= 40 + 34. 40 + ....+ 396.40 = 40.(1 + 34 + ...+ 396) chia hết cho 40
=> p chia hết cho 40
NT
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
0
8 tháng 8 2016
Bài 1
a) 34 + 35 + 36 + 37 = 34(1 + 3 + 32 + 33)\
b) a)A = 1 + 3 + 32 +......399 =(1 + 3 + 32 + 33 ) + ...+(396 + 397 + 398 + 399)
= (1 + 3 + 32 + 33 ) + .. +396(1 + 3 + 32 + 33 )
= 40 + ... + 396 . 40
= 40 (1 + 3 +...+ 396) chia hết cho 40
8 tháng 8 2016
Bài 2
a)
+)A chia hết cho 6
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{2002}\left(5+5^2\right)\)
\(A=30+5^2.30+...+5^{2002}.30\)
\(A=30\left(1+5^2+...+5^{2002}\right)\)chia hết cho 6
+)A chia hết cho 31
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{2002}+5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2+5^3\right)+5^3\left(5+5^2+5^3\right)+...+5^{2001}\left(5+5^2+5^3\right)\)
\(A=155+5^3.155+...+5^{2001}.155\)
\(A=155\left(1+5^3+...+5^{2001}\right)\)chia hết cho 31
+) A chia hết cho 156
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2000}\left(5+5^2+5^3+5^4\right)\)
\(A=780+5^4.780+...+5^{2000}.780\)
\(A=780\left(1+5^4+...+5^{2000}\right)\)chia hết cho 156
b)B=165+2^15 chia hết cho 33
ta có 165 chia hết cho 33
mà 215 ko chia hết cho 33
vậy 165+2^15 không chia hết cho 33 hay B không chia hết cho 33.
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\)
=>\(2A=1-\dfrac{1}{3^{99}}\)
=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{99}}< \dfrac{1}{2}\)