dãy số 2, 6, 12, 20...9900 tách ra thành 1.2, 2.3, 3.4, 4.5,..., 99.100 
nghĩa là mình có công thức ∑ (i=1 -> 99) (2010) / (99.(99+1)) 
(2010). ∑(i=1 -> 99) (99/100) 
2010 . (99/100) = 1989,9

đề sai viết lại đi

\(A=\frac{2010}{2}+\frac{2010}{2}+\frac{2010}{6}+\frac{2010}{12}+...+\frac{2010}{9900}\)

<=>\(A=2010\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)

<=>\(A=2010\left(\frac{1}{2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

<=>\(A=2010\left(\frac{1}{2}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

<=>\(A=2010\left(\frac{1}{2}+1-\frac{1}{100}\right)\)

<=>\(A=2010.\frac{149}{100}\)

<=>\(A=\frac{29949}{10}\)

Nếu như đề của bạn viết bị đúng thì ko sao, nhưng nếu đề bạn có bị thừa phân số 2010/2 thì chỉnh sửa lại bài làm bên trên 1 chút

\(A=\dfrac{2010}{2}+\dfrac{2010}{6}+\dfrac{2010}{12}+...+\dfrac{2010}{9900}=2010\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=2010\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2010\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2010\left(1-\dfrac{1}{100}\right)=2010.\dfrac{99}{100}=\dfrac{19899}{10}\)

\(A=\dfrac{2010}{2}+\dfrac{2010}{6}+...+\dfrac{2010}{9900}\\ =2010\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)\\ =2010\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\\ =2010\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2010.\dfrac{99}{100}=\dfrac{19899}{10}\)

pham thi ha linh sai

\(A=\frac{2010}{2}+\frac{2010}{6}+...+\frac{2010}{9900}\)

\(=2010.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)

\(=2010.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=2010.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=2010.\left(1-\frac{1}{100}\right)=2010.\frac{99}{100}\)

\(=\frac{19899}{10}\)

??????????????????????????????????????????

\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)

\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)

\(P=-3^{2010}\)