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    \(0<\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)<1\)cũng như \(\notin Z\)

Vậy B không phải là số nguyên ^_^

B  \(= \frac{2^{2} - 1}{2^{2}} + \frac{3^{2} - 1}{3^{2}} + \frac{4^{2} - 1}{4^{2}} + . . . + \frac{5 0^{2} - 1}{5 0^{2}}\)

    \(= 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)

mà    \(0 < \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) < 1\)cũng như \(\notin Z\)

Vậy B không phải là số nguyên

8/9 x 15/16 x 24/25 x...x 2499/2500

=   \(\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{4\times6}{5\times5}\times...\times\frac{49\times51}{50\times50}\)

=   \(\frac{2\times4\times3\times5\times4\times6\times...\times49\times51}{3\times3\times4\times4\times5\times5\times...\times50\times50}\)

= \(\frac{2\times51}{3\times50}\)

= \(\frac{17}{25}\)

\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{2499}{2500}=\frac{8.15...2499}{9.16...2500}=\frac{2.4.3.5...49.51}{3.3.4.4...50.50}=\frac{\left(2.3.4...49.50\right).\left(4.5...49.51\right)}{\left(2.3.4...49.50\right).\left(3.4.5...49.50\right)}=\frac{51}{3.50}\)\(=\frac{17}{50}\)

a:Sửa đề: 1x3+2x4+...+99x101

\(=1\times\left(1+2\right)+2\times\left(2+2\right)+\cdots+99\times\left(99+2\right)\)

\(=\left(1\times1+2\times2+\cdots+99\times99\right)+2\times\left(1+2+\cdots+99\right)\)

\(=\frac{99\times\left(99+1\right)\times\left(2\times99+1\right)}{6}+2\times\frac{99\times100}{2}\)

\(=\frac{99\times100\times199}{6}+99\times100=33\times50\times199+99\times100\)

\(=33\times50\times\left(199+3\times2\right)=33\times50\times205=338250\)

b: \(\frac89\times\frac{15}{16}\times\ldots\times\frac{2499}{2500}\)

\(=\left(1-\frac19\right)\times\left(1-\frac{1}{16}\right)\times\ldots\times\left(1-\frac{1}{2500}\right)\)

\(=\left(1-\frac13\right)\times\left(1-\frac14\right)\times\ldots\times\left(1-\frac{1}{50}\right)\times\left(1+\frac13\right)\times\left(1+\frac14\right)\times\ldots\times\left(1+\frac{1}{50}\right)\)

\(=\frac23\times\frac34\times\ldots\times\frac{49}{50}\times\frac43\times\frac54\times\ldots\times\frac{51}{50}=\frac{2}{50}\times\frac{51}{3}=\frac{17}{25}\)