\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=\frac11-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

1. Ba của Tý gọi bà của Tèo là Anh trai.

2. Quảng trường đỏ ở Nga.

Chihuahua

Ta có:

\(B\left(2\right)=\left\lbrace0;\pm2;\pm4;\pm6;\pm8;\ldots\right\rbrace\)

Mà x ∈ N*, x < 6

⇒ x ∈ {2; 4}

Vậy x ∈ {2; 4}

a) Ta có:

\(C=1+4+4^2+4^3+4^4+4^5+4^6\)

\(4C=\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\cdot4\)

\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)

b)

\(\Rightarrow3C=4C-C=\left(4+4^2+4^3+4^4+4^5+4^6+4^7\right)-\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\)

\(3C=4^7-1\)

Vì \(3C\vdots3\) nên \(4^7-1\vdots3\)

Vậy ....

a) Ta có:

\(A=1+3+3^2+3^3+3^4+3^5+3^6+3^7\)

\(3A=\left(1+3+3^2+3^3+3^4+3^5+3^6+3^7\right)\cdot3\)

\(3A=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\)

b)

\(\Rightarrow2A=3A-A=\left(3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\right)-\left(1+3+3^2+3^3+3^4+3^5+3^6+3^7\right)\)

\(2A=3^8-1\)

Mà \(2A\vdots2\)

\(\Rightarrow3^8-1\vdots2\)

Vậy ...

SAO V BN?

Ta có:

\(A=1+2^1+2^2+\cdots+2^{2007}\)

\(2A=2+2^2+2^3+\cdots+2^{2008}\)

\(2A-A=\left(2+2^2+2^3+\cdots+2^{2008}\right)-\left(1+2^1+2^2+\cdots+2^{2007}\right)\)

\(A=2^{2008}-1\)

Vậy:

\(a)2A=2+2^2+2^3+\cdots+2^{2008}\)

\(b)A=2^{2008}-1\)

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