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\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
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1.(2 - 1)+2.(3 - 1)+3.(4 - 1)+....+20.(20 + 1 - 1)=[(1.2 +2.3 + 3.4 +4.5 + ....+20.(20+1)] - (1 + 2 +3 + ... +20)=\(\frac{20.\left(20+1\right).\left(20.2+1\right)}{6}\)
=2870
Đặt\(A=1^2+2^2+3^2+...+20^2\)
Ta có \(A=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+3\cdot\left(4-1\right)+...+20\cdot\left(21-1\right)\)
\(A=\left(1\cdot2+2\cdot3+...+20\cdot21\right)-\left(1+2+3+...+20\right)\)
\(A=B-C\)(với \(B=\left(1\cdot2+2\cdot3+...+20\cdot21\right);C=\left(1+2+3+...+20\right)\)
Dễ nhận thấy \(C=1+2+3+...+20=\frac{20\cdot21}{2}=10\cdot21=210\)
Xét \(3B=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+20\cdot21\cdot\left(22-19\right)\)
\(3B=\left(1\cdot2\cdot3+2\cdot3\cdot4+...+20\cdot21\cdot22\right)-\left(1\cdot2\cdot3+2\cdot3\cdot4+...+19\cdot20\cdot21\right)\)
\(3B=20\cdot21\cdot22\Leftrightarrow B=20\cdot7\cdot22=3080\)
Vậy \(A=B-C=3080-210=2870\)
Nhận xét: Phương pháp giải
Tính A bằng cách đưa về những dãy số đã biết cách tính
Tính B bằng cách khử liên tiếp: số hạng sau sẽ khử số hạng liền trước.
Chúc bạn học tốt!
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\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
Sau khi lược bỏ,ta còn lại:
\(A=1-\frac{1}{20}=\frac{19}{20}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{18.19}+\frac{1}{19.20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{20}\)
\(\Rightarrow A=\frac{19}{20}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{18}-\frac{1}{19}\)
\(=1-\frac{1}{19}=\frac{18}{19}\)
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\(D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
\(E=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-1+\dfrac{1}{99}=\dfrac{2}{99}-1=-\dfrac{97}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có công thức :
Với mọi n thuộc N thì :
\(1^2+2^2+3^2+.......+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Áp dụng vào bài toán ta được :
\(A=1^2+2^2+3^2+....+20^2=\frac{20\left(20+1\right)\left(2.20+1\right)}{6}=2870\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(S=1-2+3-4+...+19-20\)
\(\Rightarrow S=\left(1-2\right)+\left(3-4\right)+...+\left(19-20\right)\) ( 10 cặp số )
\(\Rightarrow S=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\) ( 10 số )
\(\Rightarrow S=\left(-1\right).10\)
\(\Rightarrow S=-10\)
Vậy \(S=-10\)