(2n-3)2n+1 : 7 = 49n
(2n-3)2n+1 : 7 = 72n
(2n-3)2n+1 = 72n .7
(2n-3)2n+1 = 72n+1
=) 2n-3 = 7
=) n=5
vậy ...
chúc bn học tôt
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(2n-3)2n+1 : 7 = 49n
(2n-3)2n+1 : 7 = 72n
(2n-3)2n+1 = 72n .7
(2n-3)2n+1 = 72n+1
=) 2n-3 = 7
=) n=5
vậy ...
chúc bn học tôt
Chứng minh rằng:
\(\frac{1.3.5.7.9.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{1}{2^n}\)
CMR \(\frac{1.3.5.7............\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)............2n}\)=\(\frac{1}{2^n}\)
CMR : \(\frac{1.3.5.7..............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...............2n}\) =\(\frac{1}{^{2^n}}\)
CMR : A = \(\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)....\left(2n-1\right).2n}{2^n}\) là một số nguyên
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
\(1,\left(n+2\right)⋮\left(n+1\right)\)
2 ,\(8⋮\left(n-2\right)\)
3,\(\left(2n+1\right)⋮\left(6-n\right)\)
4;\(3n⋮\left(n-1\right)\)
5, \(\left(3n+5\right)⋮\left(2n+1\right)\)
6, \(\left(3n+1\right)⋮\left(2n-1\right)\)
C/m rằng B= \(\frac{1.3.5............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).......2n}=\frac{1}{2^2}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2n+1\right)\cdot\left(2n+3\right)}=\frac{n+1}{n+3}\)
\(\frac{1\cdot3\cdot5\cdot.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)=\(\frac{1}{2^n}\)với n \(\varepsilon\)N*