Question

[Nguyễn Thị Quỳnh Chi]

Cho S=\(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{43.46}\)

cmr S<1

Answer 1

[Dương Nguyễn]

Ta có:

\(S=\dfrac{3}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)

\(S=1.\left(\dfrac{1}{1}-\dfrac{1}{46}\right)\)

\(S=1.\dfrac{45}{46}=\dfrac{45}{46}\)

Vì \(\dfrac{45}{46}< \dfrac{46}{46}\) nên \(\dfrac{45}{46}< 1\).

Vậy S < 1.

Answer 2

[Phạm Thanh Hằng]

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)

\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)

Ta thấy:

\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)

\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)

\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)

\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)

\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)