Question

Cho S=3/1.4+3/4.7+3/7.10+...............+3/43.46. chứng minh rằng S<1

Answer 1

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

S<\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{43}\)-\(\dfrac{1}{46}\)

S< \(\dfrac{1}{1}\)-\(\dfrac{1}{46}\)

S<\(\dfrac{45}{46}\)<1

Vậy S< 1

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Answer 2

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{34.46}\)

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

\(S=1-\dfrac{1}{46}\)

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

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{34.46}< 1\)

\(\Rightarrow S< 1\) (đpcm)

Answer 3

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

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)

\(=1-\dfrac{1}{46}< 1\)

\(\Rightarrowđpcm\)

Answer 4

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

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

\(S=1-\dfrac{1}{46}< 1\)
=> S < 1
@Lương Thị Quỳnh Trang