Question

Chứng tỏ rằng;\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+......+\(\frac{1}{49.50}\) < 1

Answer 1

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1-\frac{1}{50}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)\)

\(=\left(1-\frac{1}{50}\right)+0+0+...+0=1-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)

Vì \(\frac{49}{50}

Answer 2

Ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

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

Mà \(\frac{49}{50}< 1\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\left(đpcm\right)\)

Answer 3

 Gọi A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

  A = 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/2 +... + 1/49 - 1/50

     = 1 - 1/50 <1

 => đpcm

Answer 4

Đây là bài toán tính tổng ta có công thức tính tổng quát: 1/n(n+1)=1/n-1/n+1suy ra A=1/1.2+1/2.3
+1/3.4+....................+1/49.50=1/1-1/2+1/2-1/3+.................+1/49-1/50=1-1/50=49/50