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Cho : B=\(\frac{4}{3}\)+\(\frac{10}{9}\)+\(\frac{28}{27}\)+...+\(\frac{3^{98}+1}{3^{98}}\)CM : B<100
\(B=\frac{3^n+1}{3^n}=1+\frac{1}{3^n}=C+D\)
B có 98 số hạng => C=98
\(D=\frac{1}{3}+\frac{..1}{3^{97}}+\frac{1}{3^{98}}\)
3.D=1+1/3+....+1/3^97
tRỪ CHO NHAU
2D=1-1/3^98
\(C=\frac{1}{2}-\frac{1}{2.3^{98}}< \frac{1}{2}\)
\(B=98+\frac{1}{2}-\frac{1}{2.3^{98}}< 99< 100\) có lẽ đề lấy 100 co chẵn. hay cộng nhầm ai tets hộ cái
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow A+2A=2^{101}-2\)
\(A\left(1+2\right)=2^{101}-2\)
\(A.3=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)
\(\Rightarrow B+3B=3^{101}-3\)
\(B\left(1+3\right)=3^{101}-3\)
\(4B=3^{101}-3\)
\(B=\frac{3^{101}-3}{4}\)
Lời giải:
a) \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)
\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow 2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
Cộng theo vế:
\(\Rightarrow B+2B=2^{201}-2\)
\(\Rightarrow B=\frac{2^{101}-2}{3}\)
c) Ta có:
\(C=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
\(\Rightarrow 3C=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
Cộng theo vế:
\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)
\(4C=3^{101}+1\Rightarrow C=\frac{3^{101}+1}{4}\)
a: \(3A=3+3^2+...+3^{101}\)
\(\Leftrightarrow2A=3^{101}-1\)
hay \(A=\dfrac{3^{101}-1}{2}\)
b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)
\(\Leftrightarrow3B=2^{101}-2\)
hay \(B=\dfrac{2^{101}-2}{3}\)
c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)
=>\(4C=3^{101}+1\)
hay \(C=\dfrac{3^{101}+1}{4}\)
Vế A
Ta có : A = 2100−299+298−297+...+22−2
2A = \(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=> 2A + A = 3A = \(2^{100}-2\Rightarrow A=\dfrac{2^{100}-2}{3}\)
=================
B làm tương tự , nhân 3 lên rồi cộng lại là ra
TA co
22b=1+1/22+1/2^4+...+1/2^96+1/2^98
b=1/2^2+1/2^4+1/2^6+.......+1/2^98+1/2^100
tu 2 dong tren tru ve theo ve TA co 3b=1-1/200
suy ra b=1/1/200 /3=1/3-1/200 /3 be hon 1/3
nen b be hon 1/3
a)M=2100-299+298-...+22-2
22M=2102-2101+2100-...+22-2
4M-M=2102-2101+2100-...+22-2-2100+299-...-22+2
3M=2102-2101
M=\(\frac{2^{102}-2^{101}}{3}\)
\(B=3^{100}-2^{100}+3^{98}-2^{98}\)
\(B=\left(3^{100}+3^{98}\right)-\left(2^{100}+2^{98}\right)\)
\(B=3^{98}\left(3^2+1\right)-2^{98}\left(2^2+1\right)\)
\(B=3^{98}.10-2^{98}.5\)
\(B=3^{98}.10-2^{97}.10\)
\(B=10\left(3^{98}-2^{97}\right)\)
Vì \(10\left(3^{98}-2^{97}\right)⋮10\)
\(\Rightarrow B⋮10\)