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A=\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{37x38x39}\)
=\(\frac{1}{2}x\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{37x38}-\frac{1}{38x39}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{38x39}\right)=\frac{185}{741}\)
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
A = 2/1x2x3 + 2/2x3x4 + 2/3x4x5 + ... + 2/36x37x38 + 2/37x38x39
A = 1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ...+ 1/36x37 - 1/37x38 + 1/37x38 - 1/38x39
A = 1/2 - 1/38x39
A = 370/741
Tớ ko chắc là đúng đâu
Trả lời:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2018.2019.2020}+\frac{1}{2.2019.2020}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2018.2019.2020}+\frac{2}{2.2019.2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}+\frac{1}{2019.2020}\right)\)
\(A=\frac{1}{2}.\frac{1}{1.2}\)
\(A=\frac{1}{4}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)
Do not ask why hay quá!
Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)
Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)
\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)
Vậy .. . .
Sửa đề chút:
\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)
\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{4}-\frac{1}{99.200}< 1\)
đpcm
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(=\frac{1}{1\times2}-\frac{1}{38\times39}=\frac{1}{2}-\frac{1}{1482}=\frac{370}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{36.37.37}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{36.37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{36.37}-\frac{1}{37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1406}\right)\)
\(=\frac{1}{2}.\frac{351}{703}\)
\(=\frac{351}{1046}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....-\frac{1}{36}+\frac{1}{37}-\frac{1}{38}+\frac{1}{38}-\frac{1}{39}\)
=\(\frac{1}{1}-\frac{1}{39}\)
=\(\frac{38}{39}\)
đảm bảo đúng
2A=2/1*2*3 +2/2*3*4 +.....+2/36*37*38+2/37*38*39
2A=1/1*2-1/2*3+1/2*3-1/3*4+....+1/37*38-1/38*39
2A=1/1*2-1/38*39
2A=1/2-1/1482
2A=370/741
A=185/741
ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương