7.
Ta có: \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{1+2+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}=2\left(\dfrac{1}{n\left(n+1\right)}\right)=2\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
Áp dụng:
\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+...+59}\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{3+1}\right)+2\left(\dfrac{1}{4}-\dfrac{1}{4+1}\right)+...+2\left(\dfrac{1}{59}-\dfrac{1}{59+1}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+2\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+...+2\left(\dfrac{1}{59}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{60}\right)\)
\(=\dfrac{19}{30}\)
6.
ta có: \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Nên: \(\dfrac{1}{n}\left(1+2+3+...+n\right)=\dfrac{1}{n}.\dfrac{n\left(n+1\right)}{2}=\dfrac{n+1}{2}\)
Áp dụng:
\(P=1+\dfrac{2+1}{2}+\dfrac{3+1}{2}+\dfrac{4+1}{2}+...+\dfrac{2012+1}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{2013}{2}\)
\(=\dfrac{2+3+4+...+2013}{2}\)
Xét tổng: \(2+3+...+2013=\left(1+2+3+...+2013\right)-1=\dfrac{2013.\left(2013+1\right)}{2}-1=2027090\)
\(\Rightarrow P=\dfrac{2027090}{2}=1013545\)
giúp em vớii :33 e cảm ơn ạ ><
