\(\frac{1}{n}-\frac{1}{n+1}=\frac{1.n+1}{n.\left(n+1\right)}-\frac{1.n}{\left(n+1\right).n}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}=\frac{n+1-n}{n.\left(n+1\right)}=\frac{1}{n.\left(n+1\right)}=\frac{1.1}{n.\left(n+1\right)}\)
\(=\frac{1}{n}.\frac{1}{n+1}\)
Chứng minh rằng :
a, \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)
b, \(\frac{1}{n\left(n+q\right)}=\frac{1}{q}\left(\frac{1}{n}-\frac{1}{n+q}\right)\)
Chứng minh rằng : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
với n thuộc N* hãy chứng tỏ rằng :
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Chứng minh rằng:
\(\frac{1}{n}.\frac{1}{\left(n+1\right)}=\frac{1}{n}-\frac{1}{\left(n+1\right)}\)
chứng tỏ rằng với mọi n thuộc N* ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
Chứng minh rằng:
a)\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)với n thuộc N*
Chứng minh rằng:
\(\frac{1.3.5.7.9.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{1}{2^n}\)
\(\frac{1}{n.\left(n+1\right).\left(n+2\right)}=\frac{1}{2}.\left[\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right]\)
Ai làm được mk sẽ tick
Chứng minh rằng:
a,\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b,\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)
Biết rằng n thuộc N*
