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Bài 1
A= 3.4 + 4.5+ 5.6+ .......+ 58.59 + 69.60
3A = 3.4.3 + 4.5.3+ 5.6.3+ .......+ 58.59.3 +59.60.3
= 3.4.(5-2) + 4.5.(6-3)+ 5.6.(7-4)+ .......+ 58.59.(60-57) +59.60.(61-58)
= 3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+..........+ 58.59.60-57.58.59+ 59.60.61-58.59.60
=2.3.4+ 59.60.61= 215964
A= 215964: 3= 71988
Bài 2:
A = 2.4 +4.6+ 6.8+.........+ 96.98+98.100
6A= 2.4.6+4.6.6+6.8.6+.........+96.98.6+98.100.6
= 2.4.6+ 4.6.(8-2) +6.8.(10-4)+.........+96.98.( 100-94) + 98 .100.( 102 - 96)
= 2.4.6+4.6.8-2.4.6 + 6.8.10 -4.6.8+..........+ 96.98.100-94.96.98+ 98.100.102-96.98.100
= 98 .100 .102= 999600
A= 999600:6= 166600
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a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1\times2\times3+2\times4\times6+4\times8\times12+7\times14\times21}{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}\\ =\dfrac{1\times2\times3}{1\times3\times5}+\dfrac{2\times4\times6}{2\times6\times10}+\dfrac{4\times8\times12}{4\times12\times20}+\dfrac{7\times14\times21}{7\times21\times35}\\ =\dfrac{2}{5}+\dfrac{4}{10}+\dfrac{8}{20}+\dfrac{14}{35}\\ =\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}\\ =\dfrac{2}{5}\times4\\ =\dfrac{8}{5}\)
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\(=\frac{\left(3.4.12^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.12^{32}}{11.2^{13}.2^{22}-2^{26}}\)
\(=\frac{3^2.2^4.2^{64}.3^{32}}{11.2^{35}-2^{26}}\)
\(=\frac{3^{34}.2^{68}}{2^{26}\left(11.2^9-1\right)}\)
\(=\frac{3^{34}.2^{42}}{11.2^9-1}\)
Còn lại làm típ nha :)