a=1.2+2.3+3.4+4.+....+200.201
các bạn tính giúp mình với
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cách mình đúng;
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n +1)3
= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...+ n(n + 1)((n + 2) - (n -1))
= 1.2.3 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)
= n(n + 1)(n + 2)
=> S = n(n + 1)(n + 2)/3
3A=1.2.3+2.3.3+3.4.3+...+19.20.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
3A=19.20.21
=> \(A=\frac{19.20.21}{3}=2660\)
mk dùng cách của lớp 8 nha bạn ;
ta có công thức xích ma như sau x(x+1)
nhập vào xích ma ta có kết quả 2660
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=9\left(1-\frac{1}{2020}\right)\)
\(=9.\frac{2019}{2020}\)
\(=\frac{18171}{2020}\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(A=9\left(1-\frac{1}{2020}\right)=\frac{9.2019}{2020}=\frac{18171}{2020}\)
...
Ta có : A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
\(\Rightarrow\)3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2).....n.(n+1).[(n+2)-(n-1)]
\(\Rightarrow\)3A= 1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+n.(n+1)(n+2)-(n-1)n(n+1)
\(\Rightarrow\)3A= (1.2.3-1.2.3)+(2.3.4-2.3.4)+....+[(n-1).n.(n+1)-(n-1)n(n+1)]+n.(n+1)(n+2)
\(\Rightarrow\)3A=n.(n+1)(n+2)
\(\Rightarrow\)A=\(\frac{\text{n.(n+1)(n+2)}}{3}\)
A = 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = 33.100.101
A = 333300
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{2014}{2015}\)
\(\Leftrightarrow A=\frac{2014}{2015}\div\frac{1}{4}\)
\(\Leftrightarrow A=\frac{8056}{2015}\)
Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51
=> 3A = 49.50.51
= >A = 49.50.51/3 = 41650
a=1.2+2.3+3.4+4.+....+200.201
3A = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 200.201.(202 - 199)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + .... + 200.201.202
3A = 200.201 . 202
A = 2706800
\(A=1.2+2.3+3.4+...+200.201\)
\(\frac{1}{A}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{201}=\frac{200}{201}\)
\(A=1:\frac{200}{201}=\frac{1.201}{200}=\frac{201}{200}\)