Đề bài yêu cầu làm j z bn

\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{49.51}{50^2}\)

\(=\frac{2.3....49}{3.4....50}.\frac{4.5....51}{3.4....50}\)

\(=\frac{2}{50}.\frac{17}{1}\)

\(=\frac{17}{25}\)

Ta có : \(A=\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)

\(A=\frac{8.15.....2499}{9.16.....2500}\)

\(A=\frac{\left(2.4\right).\left(3.5\right).....\left(49.51\right)}{\left(3.3\right).\left(4.4\right).....\left(50.50\right)}\)

\(A=\frac{\left(2.3....49\right).\left(4.5....51\right)}{\left(3.4....50\right).\left(3.4.....50\right)}\)

\(A=\frac{2\left(3.4.....49\right).\left(4.5.....50\right).51}{\left(3.4.....49\right).50.3.\left(4.5.....50\right)}\)

\(A=\frac{2.51}{3.50}\)

\(A=\frac{2.17.3}{3.25.2}\)

\(A=\frac{17}{25}\)

So thú bi cháy con gì ra đau tiên:

Do S = \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}\)

\(\Rightarrow\)S = \(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=(1+1+1+...+1) - \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=49-\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Dễ thấy:\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)không phải là số tự nhiên

\(\Rightarrow\)S\(\notin N\)

17/25 nha bạn

B  \(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{50^2-1}{50^2}\)

    \(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)

mà    \(0

mình cũng đang định hỏi giống bạn !!!

hảo năm 2018 mới trả lời =)