Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)\)

\(A=1-\frac{1}{2^{2013}}\)

\(A=\frac{2^{2013}-1}{2^{2013}}\)

Vậy \(A=\frac{2^{2013}-1}{2^{2013}}\)

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Giúp mình câu sau với

\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2014}}\)

\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)

\(3A-A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2014}}\right)\)

\(2A=\frac{1}{3}-\frac{1}{3^{2014}}\)

\(A=\frac{\frac{1}{3}-\frac{1}{3^{2014}}}{2}\)

\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(A=1-\frac{1}{2^9}\)

\(A=\frac{2^9-1}{2^9}\)

Vậy \(A=\frac{2^9-1}{2^9}\)

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Em cứ lấy máy tính bấm bài 1 đi là đc

Câu 1:

A = -3/12 + 5/7 - (-1)/42

A = -21/84 + 60/84 + 2/84

A = 39/84 + 2/84

A = 41/84

A = 1/3^2 + 1/4^2 + 1/5^2 + ... + 1/50^2

1/3^2 = 1/9

1/4^2 < 1/3.4 = 1/3 - 1/4

1/5^2 < 1/4.5 = 1/4 - 1/5

.............................................

1/50^2 < 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+..+1/50^2 = 1/9 + 1/3 - 1/50

A = 4/9 - 1/50 < 4/9

1/3^2 = 1/9

1/4^2 > 1/4.5 = 1/4 - 1/5

1/5^2 > 1/5.6 = 1/5 - 1/6

............................................

1/50^2 > 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+ ...+ 1/50^2 > 1/9+1/4-1/50

A > 1/4 + (1/9 - 1/50)

1/9 > 1/50

1/9 - 1/50 > 0

A > 1/4 + 1/9 - 1/50 > 1/4

Vậy 1/4 < A < 4/9 (đpcm)

Câu a:

33 ⋮ (x+ 1)

(x+ 1) ∈ Ư(33) = {-33; -11; -3; -1; 1; 3; 11}

x ∈ {-34; -12; -4; -2; 0; 2; 10}

Vậy: {-34; -12; -4; -2; 0; 2; 10}

Câu b:

x ∈ ƯC(250; 48)

250 = 2.5^3; 48 = 2^4.3

ƯCLN(250; 48) = 2

x ∈ ƯC(2) = {-2; -1}

Vậy x ∈ {-2; -1}