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

S=43​+98​+...+25002499​

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

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

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

\(\frac{1}{2^{2}} < \frac{1}{1 \cdot 2} = 1 - \frac{1}{2}\)

\(\frac{1}{3^{2}} < \frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}\)

...

\(\frac{1}{5 0^{2}} < \frac{1}{49 \cdot 50} = \frac{1}{49} - \frac{1}{50}\)

Do đó: \(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50}\)

=>\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)

=>\(0 < \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)

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

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

=>49>B>48

=>S không là số tự nhiên

S=43​+98​+1615​+...+50002499​

\(S = 1 - \frac{1}{4} + 1 - \frac{1}{9} + 1 - \frac{1}{16} + . . . + 1 - \frac{1}{5000}\)

\(S = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{4} + + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{5000} \left.\right)\)

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

Lại có : 

\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{49.50}\)

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

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

\(\Rightarrow\)\(S = 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > 49 - 1 = 48\)\(\left(\right. 2 \left.\right)\)

Từ (1) và (2) suy ra : 

\(48 < S < 49\)

Vậy S không là số tự nhiên 

Chúc các bạn học tốt nhé ! =))

\(=\frac{2\cdot4}{3^2}\cdot\frac{3.5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot......\cdot\frac{49\cdot51}{50^2}\)

=\(\frac{\left[2\cdot3\cdot4\cdot......\cdot49\right]\cdot\left[4\cdot5\cdot6\cdot.....\cdot51\right]}{\left[3\cdot4\cdot5\cdot....\cdot50\right]\cdot\left[3\cdot4\cdot5\cdot....\cdot50\right]}\)

=\(\frac{2\cdot51}{50\cdot3}\)

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

Vì \(\frac{17}{25}\) ko phải là số nguyên nên B ko phải là số nguyên [ĐPCM]

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

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

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

Mặt khác ta có:

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< 1-\frac{1}{50}< 1\)

\(\Rightarrow B=49-A>49-1=48\)

\(\Rightarrow48< B< 49\)

\(\Rightarrow\) B nằm giữa 2 số nguyên liên tiếp nên B không phải là số nguyên

\(B=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+\frac{1}{2500}\)

\(B=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+\frac{1}{50^2}=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)\)(từ 2 đến 50 có 49 số nên có 49 số 1)

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

Nhận xét: \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{50^2}<\frac{1}{49.50}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}<1\) => \(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)>-1\)

=> \(B=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)>49-1=48\)(2)

Từ (1)(2) => 48 < B < 49 => B không phải là số nguyêm