\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}\)Tính tổng sau
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Tính tổng 10 phân số đầu tiên của dãy sau :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}=?\)
Tính nhanh lên mình đang cần gấp
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
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bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
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\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)\(=\frac{6}{12}-\frac{1}{12}=\frac{5}{12}\)
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H24
26 tháng 2 2018 lúc 13:36
a, Chứng tỏ rằng với mọi \(n\inℕ^∗\) ta luôn có: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
b, Áp dụng tính nhanh tổng sau: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+....+\frac{1}{90}\)
a ) Ta có : \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\) \(=\frac{1}{n.\left(n+1\right)}\)
b ) Áp dụng công thức trên tính tổng này như sau :
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chúc học giỏi !!!
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a, \(VP=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}\)
\(=\frac{1}{n\left(n+1\right)}=VT\RightarrowĐPCM\)
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a) Ta có: \(VP:\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}=VT\)
v) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
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tính A = \(\frac{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}}{5+\frac{5}{3}+\frac{5}{6}+\frac{1}{2}+...+\frac{1}{9}}\)
tính A = \(\frac{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}}{5+\frac{5}{3}+\frac{5}{6}+\frac{1}{2}+...+\frac{1}{9}}\)
tks
Tính tổng :
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(\frac{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}}{5+\frac{5}{3}+\frac{5}{6}+\frac{1}{2}+...+\frac{1}{9}}\)
Tính nhanh. M.n giúp dùm nhak. Tks
00
26 tháng 6 2017 lúc 13:01
Tính thuận tiện :
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
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\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\)\(\frac{1}{5}\)\(+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}+\frac{1}{10x11}\)
= \(\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{10-9}{9x10}+\frac{11-10}{10x11}\)
= \(\frac{2}{1x2}-\frac{1}{1x2}+\frac{3}{2x3}-\frac{2}{2x3}+...+\frac{10}{9x10}-\frac{9}{9x10}+\frac{11}{10x11}-\frac{10}{10x11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}=\frac{10}{11}\)
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Tính
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{52}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
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\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}.\)
tính A
Nhớ rằng: \(\frac{n}{a.\left(a+n\right)}=\frac{1}{a}+\frac{1}{a+n}\)
Ta có: \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Áp dụng công thức trên, ta có:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
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