Tính A=1/3.6+1/6.9+.....+1/113.116
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\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
\(S=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{219.222}\)
\(\Rightarrow3S=\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{219.222}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{219}-\frac{1}{222}\)
\(=\frac{1}{3}-\frac{1}{222}< \frac{1}{3}\)
\(\Rightarrow S< \frac{1}{9}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
E=1/1.3+1/3.6+1/6.9+.............+1/20.23
<=>E=1/1-1/3+1/3-1/6+1/6-1/9+...........+1/20-1/23
<=>E=1/1-1/23
<=>E=22/23
Kb và k mk nha mn.
Cho hai phan so 1/n va 1/n+1 (n thuoc z)chung to rang h cua hai phan so nay bang hieu cua chung
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)