Question

chứng tỏ rằng:

1/1.2+1/2.3+1/3.4+...+1/49.50<1

Answer 1

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

Ta có: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)

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

= \(\dfrac{1}{1}-\dfrac{1}{50}< 1\)

= \(\dfrac{50}{50}+\dfrac{-1}{50}< 1\)

= \(\dfrac{49}{50}< 1\)

Vậy \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)

Answer 2

1/1.2 = 2 đã lớn hơn 1 rồi @@