Tính tổng:
A = 1.2 + 2.3 + 3.4 + ... + 19.20
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\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=9\left(1-\frac{1}{2020}\right)\)
\(=9.\frac{2019}{2020}\)
\(=\frac{18171}{2020}\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(A=9\left(1-\frac{1}{2020}\right)=\frac{9.2019}{2020}=\frac{18171}{2020}\)
...
Ta có : \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)
= 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)
= 1 - \(\frac{1}{7}\)= \(\frac{6}{7}\)
\(S=9,8+8,7+7,6+...+2,1-1,2-2,3-3,4-...-8,9\)
\(=\left(9,8-8,9\right)+\left(8,7-7,8\right)+...+\left(2,1-1,2\right)\)
\(=0,9+0,9+...+0,9\)
\(=0,9\times8=7,2\)
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
\(A=1.2+2.3+3.4+...+200.201\)
\(\frac{1}{A}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{201}=\frac{200}{201}\)
\(A=1:\frac{200}{201}=\frac{1.201}{200}=\frac{201}{200}\)
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
S = 1.2 + 2.3 + 3.4 + ... + 99.100
3S = 1.2.3 + 2.3.3 + 3.4.3 + .. +99.100.3
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3S = 99.100.101
3S = 999 900
S = 333 300
Vậy S = 333 300
\(A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 100 . 101\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+100\cdot101\cdot\left(102-99\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+100\cdot101\cdot102-99\cdot100\cdot101\)
\(3A=100\cdot101\cdot102\)
\(3A=1030200\)
\(A=1030200\text{ : }3\)
\(A=343400\)
\(A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 100 . 101\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+100\cdot101\cdot\left(102-99\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+100\cdot101\cdot102-99\cdot100\cdot101\)
\(3A=100\cdot101\cdot102\)
\(3A=1030200\)
\(A=1030200\text{ : }3\)
\(A=343400\)
Câu hỏi của Nguyen Yen Nhi - Toán lớp 6 - Học toán với OnlineMath
Em xem bài ở link này!
\(S=1.2+2.3+3.4+...+99.100\)
\(\Leftrightarrow3S=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(\Leftrightarrow3S=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+99.100\left(101-98\right)\)
\(3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3S=99.100.101\)
\(\Leftrightarrow S=\frac{99.100.101}{3}\)
\(\Leftrightarrow S=33.100.101\)
\(\Leftrightarrow S=333300\)
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