tính : 1+1\2*(1+2)+1\3*(1+2+3)+...+1\6*(1+2+3+...+16)
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b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)
\(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)
\(=1+1,5+2+...+8,5\)
\(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)
B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)
Vậy 1/8<2 hay 1/8<16/8
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a)\(=\dfrac{16}{13}-\dfrac{3}{15}+\dfrac{6}{13}=\dfrac{22}{13}-\dfrac{3}{15}=\dfrac{96}{65}\)
b)\(=\dfrac{21}{8}-\left(\dfrac{5}{10}+\dfrac{6}{10}\right)=\dfrac{21}{8}-\dfrac{11}{10}=\dfrac{61}{40}\)
c)\(=\dfrac{27}{10}-3-\dfrac{4}{7}--\dfrac{61}{70}\)
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a) \(\frac{-3}{16}+\frac{-7}{16}+\frac{6}{-16}\)
\(=-\frac{3}{16}+-\frac{7}{16}+\frac{-6}{16}\)
\(=\frac{\left(-3\right)+\left(-7\right)+\left(-6\right)}{16}\)
\(=\frac{-16}{16}=-1\)
b) \(\frac{2}{5}-\frac{1}{7}+\frac{7}{10}\)
\(=\frac{28}{70}-\frac{10}{70}+\frac{49}{70}\)
\(=\frac{28-10+49}{70}\)
\(=\frac{67}{70}\)
c) \(3+\frac{1}{3}-2\frac{1}{2}\)
\(=\frac{3}{1}+\frac{1}{3}-\frac{5}{2}\)
\(=\frac{18}{6}+\frac{2}{6}-\frac{15}{6}\)
\(=\frac{18+2-15}{6}\)
\(=\frac{5}{6}\)
d) \(-\frac{3}{5}-\left(-\frac{2}{15}\right)\)
\(=-\frac{3}{5}+\frac{2}{15}\)
\(=\frac{-9}{15}+\frac{2}{15}\)
\(=-\frac{7}{15}\)
\(\frac{-3}{16}\)+(\(\frac{-7}{16}\) ) +(\(\frac{6}{-16}\))
=\(\frac{-3-7-6}{16}\)
=\(\frac{-16}{16}\)= \(-1\)
\(\frac{2}{5}\)\(-\)\(\frac{1}{7}\)+\(\frac{7}{10}\)
=\(\frac{9}{35}\)+\(\frac{7}{10}\)
=\(\frac{67}{70}\)
3+ \(\frac{1}{3}\)\(-\) \(\frac{21}{2}\)
=\(\frac{18+2-63}{6}\)
=\(\frac{-43}{6}\)
\(\frac{-3}{5}\)\(-\)(\(\frac{-2}{15}\))
=\(\frac{-3}{5}\)+\(\frac{2}{15}\)
=\(\frac{-9+2}{15}\)
=\(\frac{-7}{15}\)
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\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ \Rightarrow A=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ \Rightarrow A=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}+\dfrac{17}{2}\\ \Rightarrow A=\dfrac{1}{2}\left(2+3+4+...+17\right)=76\)
Cho n\(\in\)N*.CMR:
\(\frac{1}{n}.\left(1+...+n\right)=\frac{n+1}{2}\)
Ta có công thức:1+2+3+.....+n=\(\frac{n.\left(n+1\right)}{2}\)
Thật vậy:\(\frac{1}{n}.\left(1+2+.....+n\right)=\frac{n+1}{2}\)
\(\Rightarrow\frac{1}{n}.\frac{n.\left(n+1\right)}{2}=\frac{n+1}{2}\)
\(\Rightarrow\frac{n.\left(n+1\right)}{n.2}=\frac{n+1}{2}\)
\(\Rightarrow\frac{n+1}{2}=\frac{n+1}{2}\)(đúng)
Thay vào ta có:\(1+\frac{1}{2}.\left(1+2\right)+.......+\frac{1}{16}.\left(1+2+3+....+16\right)\)
=\(1+\frac{2+1}{2}+.....+\frac{16+1}{2}\)
=\(1+\frac{3}{2}+.......+\frac{17}{2}\)
=\(\frac{2+3+....+17}{2}\)
=\(\frac{152}{2}\)
=76
\(A=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+...+\frac{1}{16}\cdot\left(1+2+3+...+16\right).\)
Tổng của n số tự nhiên liên tiếp là: \(1+2+3+...+n=\frac{n\cdot\left(n+1\right)}{2}\).
\(A=\frac{2}{2}+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{16}\cdot\frac{16\cdot17}{2}.\)
\(=-\frac{1}{2}+\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=-\frac{1}{2}+\frac{1+2+3+4+...+17}{2}=-\frac{1}{2}+\frac{\frac{17\cdot18}{2}}{2}=-\frac{1}{2}+\frac{153}{2}=\frac{152}{2}=76\)
Đ/S A = 76