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Ta thấy
1.3.5.....19 = 1.2.3.4.5.6.....19.20/2.4.6.....20
= 1.2.3.4.5.6.....19.20/1.2.2.2.3.2.....10.2
= 1.2.3.4.5.6.....19.20/(1.2.3.....10).(2.2.2.....2)
= (1.2.3.4.5.6.....10).(11.12.13.....19.10)/(1.2.3.....10).(2.2.2.....2)
= 11.12.13.....19.20/2.2.2.....2
= 11/2 . 12/2 . 13/2 . ... . 19/2 . 20/2
=> Đpcm
Ta có:
1.3.5...19 = 1.2.3.4.5.6...19.20/2.4.6...20
= 1.2.3.4.5.6...19.20/2^10.(1.2.3...10)
= 11.12.13....20/2^10
= 11/2 . 12/2 . 13/2 ... 20/2 ( đpcm)
ta có : 1.3.5.....19. (2.4.6...20/2.4.6....20)=1.2.3....19.20/(1.2.3....10)(2.2.2.2.2.2.2.2.2.2)=11/2 .12/2 ......20/2
10 chữ số 2
Lời giải:
\(\frac{11}{2}.\frac{12}{2}.\frac{13}{2}....\frac{20}{2}=\frac{(12.14.16.18.20)(11.13.15.17.19)}{2^{10}}\)
\(=\frac{12}{2^2}.\frac{14}{2}.\frac{16}{2^4}.\frac{18}{2}.\frac{20}{2^2}(11.13.15.17.19)\)
\(=3.7.1.9.5.11.13.15.17.19=1.3.5.7.9.11.13.15.17.19\)
Ta có:
\(1.3.5.7.9...\left(2n-1\right)=\frac{\left[1.3.5.7.9....\left(2n-1\right)\right].\left[2.4.6.8...2n\right]}{2.4.6.8....2n}=\frac{1.2.3.4.5.6....2n}{\left(2.1\right).\left(2.2\right).\left(2.3\right)\left(2.4\right)....\left(2.n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{1.2.3.4.5.6....2n}{\left(2.2.2.....2\right).\left(1.2.3.4.....n\right)}=\frac{\left(1.2.3.4.....n\right)\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}{2^n.\left(1.2.3.4....n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n}\)
=> \(\frac{1.3.5.7.9...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}=\frac{1}{2^n}\)(đpcm)
ai nhanh mik cho love you my friend best is my friend