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S<1/1.2+1/2.3+...+1/14.15=1-1/15=14/15=>S<14/15(*)
S>1/1.2.3+1/2.3.4+...+1/14.15.16=1/2(1/2-1/15.16)=119/480>7/16=>S>7/16(**)
Từ * và ** suy ra đpcm
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{15^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{15^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{14.15}\)
Ta có:
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}=\frac{3-2}{3.2}+\frac{4-3}{4.3}+...+\frac{16-15}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{14.15}=\frac{2-1}{1.2}+\frac{3-2}{3.2}+...+\frac{15-14}{15.14}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{14}-\frac{1}{15}=1-\frac{1}{15}=\frac{14}{15}\)
Vậy \(\frac{7}{16}< S< \frac{14}{15}\)
Ta có : \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
Ta có:
\(S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Bài toán phụ 1:
Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12x3=1/4 (1)
Bài toán phụ 2:
Ta có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60x3=1/20 (2)
Từ (1) và (2), ta có:
1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<4/20+5/20+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2
=>1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
a) Ta có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}>\frac{6}{19}.5=\frac{30}{19}>1\)
\(\Rightarrow S>1\)
Ta lại có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}< \frac{6}{15}.5=\frac{30}{15}=2\)
\(\Rightarrow S< 2\)
Vậy, 1 < S < 2
b) \(1< S< 2\Rightarrow S\notin Z\)
a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy...
b, Đặt A là tên của tổng trên
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B là biêu thức trong ngoặc
Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)
Thay B vào A ta được:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)