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a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
Bài 4:
=>(x-5)*3/10=1/5x+5
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=6,5
=>0,1x=6,5
=>x=65
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)\)
\(=\dfrac{1}{3}.\dfrac{3n}{6n+4}\)
\(=\dfrac{n}{6n+4}\) ( đpcm )
Vậy...
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)(đpcm)
\(A=\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{n+4}\right)\\ =\dfrac{1}{2}.\dfrac{n+1}{3\left(n+4\right)}=\dfrac{n+1}{6\left(n+4\right)}\\ =\dfrac{n+4-3}{6\left(n+4\right)}=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}< \dfrac{1}{6}.\)
Giải:
A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)
A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)
A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)
A=1/2.(1/3-1/n+4)
A=1/6-1/2.(n+4)
⇒A>1/6
Chúc bạn học tốt!
\(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)
\(\dfrac{a}{n+\left(n+a\right)}+\dfrac{1}{n+a}=\dfrac{1}{n}\)
Vậy ta sẽ CRM\(\dfrac{a}{n+\left(n+a\right)}+\dfrac{1}{n+a}=\dfrac{1}{n}\)
\(\dfrac{a}{n\left(n+a\right)}+\dfrac{1}{n+a}\)
\(=\dfrac{a}{n}\cdot\dfrac{1}{\left(n+a\right)}+\dfrac{1}{n+a}\)
\(=\dfrac{1}{n+a}\cdot\left(\dfrac{a}{n}+1\right)\)
\(=\dfrac{1}{n+a}\cdot\dfrac{a+n}{n}\)
Đã \(CMR:\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)
\(\dfrac{1}{n}-\dfrac{1}{n+k}=\dfrac{n+k}{n\left(n+k\right)}-\dfrac{n}{n\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{k}{n\left(n+k\right)}\)
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)(đpcm)
\(VP=\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}=VT\)
Ta có :
\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}hay\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)