Tính A=1.2.3+2.3.4+,,,,,,+98.99.100
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A = 1.2.3 + 2.3.4 + …. + 98.99.100
4A = 4( 1.2.3 + 2.3.4 + …. + 98.99.100)
4A= 1.2.3.4 + 2.3.4.4 +....+98.99.100.4
4A= 1.2.3.4 + 2.3.4 (5-1) +....+98.99.100(101- 97)
4A= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ....+ 98.99.100.101 - 97.98.99.100
4A= 98.99.100.101
4A=97990200
A= 97990200:4
A=24497550
Vậy.....
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
Đặt S=1.2.3+2.3.4+...+98.99.100
=>4S=1.2.3.4+2.3.4.4+...+98.99.100.4
=>3S=1.2.3(4-0)+2.3.4(5-1)+....+98.99.100(101-97)
=>4S=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+....+98.99.100.101-97.98.99.100
=>4S=98.99.100.101
=>S=24497550
Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3};\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4};...;\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Qua công thức trên, bạn có thể rút ra tổng quát: (đây là mình nói thêm)
\(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n-2\right)}=\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
Ta suy ra:
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
Thấy \(-\frac{1}{2.3}+\frac{1}{2.3}=0;-\frac{1}{3.4}+\frac{1}{3.4}=0;...\)
\(\Rightarrow2B=\frac{1}{2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Mình nhầm, công thức tổng quát mình nói thêm bạn đổi cái n-2 thành n+2 nha
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
=>A=98.99.100.101/4
Câu hỏi của hồ thị hằng - Toán lớp 6 - Học toán với OnlineMath
\(4D=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+.....+98.99.100\left(101-97\right)\)
\(\Rightarrow4D=1.2.3.4+2.3.4.5-1.2.3.4+.....+98.99.100.101-97.98.99.100\)
\(\Rightarrow4D=98.99.100.101\)
\(\Rightarrow D=\frac{98.99.100.101}{4}\)
\(\Rightarrow D=24497550\)
A= 1.2.3 + 2.3.4 + 3.4.5 +.....+ 98.99.100
4A = 98.99.100.4 + .....+ 3.4.5.4 + 2.3.4.4 + 1.2.3.4
4A = 98.99.100.(101-97) +... + 2.3.4.(5-1) + 1.2.3.4
4A = 98.99.100.101 - 97.98.99.100+......+2.3.4.5 - 1.2.3.4 + 1.2.3.4
4A = 98.99.100.101
A = 98.99.100.101 : 4
A = 24497550
Ta có: A=1.2.3+2.3.4+…+98.99.100
=>A.4=1.2.3.4+2.3.4.4+…+98.99.100.4
=>A.4=1.2.3.(4-0)+2.3.4.(5-1)+…+98.99.100.(101-97)
=>A.4=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+…+98.99.100.101-97.98.99.100
=>A.4=98.99.100.101
=>A.4=97990200
=>A=97990200:4
=>A=24497550
Michiel Girl Mít ướt bài lớp 7 đó, lm đi