1 mũ 2+ 2 mũ 2+ 3 mũ 2+ ...+99 mũ 2+ 100 mũ 2
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A=1 mũ 2+2 mũ 2+3 mũ 2 +...+100 mũ 2
B=1 mũ 2 +3 mũ 2+5 mũ 2+...+99 mũ 2
C=2 mũ 2+4 mũ 2+...+20 mũ 2
D=1 mũ 2+4 mũ 2+7 mũ 2+...+100 mũ 2
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A=2+2 mũ 2+2 mũ 3+.....+2 mũ 99+2 mũ 100
B=1+5+5 mũ 2+5 mũ 3+......+5 mũ 99+5 mũ 100
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A = 2 + 22 + 23 +...+299 + 2100
2A = 22 + 23 +...+ 299 + 2100 + 2101
2A - A = 2101 - 2
A = 2101 - 2
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B = 1 + 5 + 52 + 53 +...+ 599 + 5100
5B = 5 + 52 + 53 +...+ 599 + 5100 + 5101
5B - B = 5101 - 1
4B = 5101 - 1
B = \(\dfrac{5^{101}-1}{4}\)
B =
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CMR: A=1/2 - 2/2 mũ 2 + 3/2 mũ 3 - ........+ 99/2 mũ 99- 100/2 mũ 100 < 2/9
1/3 - 2/3 mũ 2 + 3/3 mũ 3 - 4/3 mũ 4 + 5/3 mũ 5 - ... + 99/3 mũ 99 - 100/3 mũ 100 < 3/16
gọi biểu thức trên là A , ta có :
\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
từ đó ta được :
\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)
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Tính :
C =(-1) mũ 2 .( -2 ) mũ 1 + (-1) mũ 3 . ( -2 ) mũ 2 + (-1 ) mũ 4 .(-2) mũ 3 + ...+( -1) mũ 100.(-2 )mũ 99
100 mũ 2-99 mũ 2+98 mũ 2-...-3 mũ 2+2 mũ 2-1 mũ 2
Gọi tổng trên là S
\(S=100^2-99^2-98^2-....-1=100^2-\left(100-1\right)^2-\left(100-2\right)^2-.....-\left(100-99\right)^2=100^2-100^2-100^2-.....-100^2+2.100+2.2.100+2.3.100+.....+2.99.100-1^2-2^2-3^2-....-99^2-100^2+100^2\)
\(A=1^2+2^2+99^2+100^2\)
=1.(2-1)+2.(3-1)+3.(4-1)+....+99.(100-1)+100.(101-1)
=1.2-1.1+2.3-1.2+3.4-1.3+...+99.100-1.99+100.101-1.100
=(1.2+2.3+3.4+...+99.100+100.101)-(1+2+3+...+100)
= [1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99) ] /3 + [(100+1).100 /2]
=[1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+100.101.102-99.10.101]/3 + 5050
=100.101.102/3 + 5050
=348450
\(\Rightarrow S=-99.100^2+2.100.99.100-A=641550\)
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(2 mũ 99 . 28-2 mũ 99 .3 mũ 3 +2mux 99 .5 mũ 2).2 mũ 100
1/2 + 1/2 mũ 2 + 1,2 mũ 3 + 1,2 mũ 4 + 3 chấm ba chấm + 1,2 mũ 99 + 1/2 mũ 100
2A = 1 + \(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+...+\(\dfrac{1}{2^{99}}\)
2A - A= 1- \(\dfrac{1}{2^{100}}\)
A= 1
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B=1/3+2/3 mũ 2+3/3 mũ 3+4/3 mũ 4+...+99/3 mũ 99+100/ 3 mũ 100 chứng minh B < 3/16
