\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}< 1\)
\(\left(\frac{3}{a\cdot\left(a+3\right)}=\frac{a+3-3}{a\cdot\left(a+3\right)}=\frac{1}{a}-\frac{1}{a+3}\right)\)
\(S=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{43\times46}\)
\(3S=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...+\frac{3}{43}-\frac{3}{46}\)
\(3S=3-\frac{3}{46}\)
\(3S=\frac{135}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy ra có điều phải chứng minh
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy S < 1 (đpcm)
s=\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}\)
\(\frac{3}{1\cdot4}=1-\frac{1}{4}\)
\(\frac{3}{4\cdot7}=\frac{1}{4}-\frac{1}{7}\)
................................
\(\frac{3}{43\cdot46}=\frac{1}{43}-\frac{1}{46}\)
=> S=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{47}\)
=>S=1-\(\frac{1}{47}\)
=> S= \(\frac{46}{47}\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{46}{46}-\frac{1}{46}\)
\(S=\frac{45}{46}\)
=> S < 1 ( vì tử số nhỏ hơn mẫu số )
Ta có \(S\) \(=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+.....+\frac{3}{43\cdot46}\)\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{43}-\frac{1}{46}\)
\(\Rightarrow S=1-\frac{1}{46}< 1\)
Vậy S < 1
k mk nha bạn.mk nhanh nhất hoặc 2
